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Here's a proof that if $X$ is any simply connected Hausdorff space such that $X\setminus \{p\}$ is path connected for all $p\in X$, then the complement of any totally disconnected subset is connected. In particular, if $X$ has more than one point then it cannot be the disjoint union of two totally disconnected subsets. Taking $X=\mathbb{R}^2$ answers the question.

[This is inspired by Włodzimierz Holsztyński's first answer, by trying to prove the result in a similar way, but using as few advanced properties of $\mathbb{R}^2$ as possible. I expect that the connectedness properties can be replaced by a statement about the Čech cohomology, such as $\check{H}^2(X)=0$ and $X\setminus\{p\}$ is connected for each $p$].

I will use proof by contradiction. So, suppose that $A\subseteq X$ is totally disconnected such that $B\equiv X\setminus A$ is not connected. Then, there are open $U,V\subseteq X$ such that $B\cap U$, $B\cap V$ are disjoint and nonempty and such that $B\subseteq U\cup V$. So, $W= U\cap V$ is disjoint from $B$ and, hence, $W\subseteq A$. If $W$ was nonempty, then choosing points $p\in W$ and $q\in X\setminus\{p\}$, there is a continuous $\gamma\colon[0,1]\to X$ with $\gamma(0)=p$, $\gamma(1)=q$. Letting $t$ be maximal such that $[0,t)\subseteq\gamma^{-1}(W)$ then $t > 0$ and the Hausdorff hypothesis implies that $\gamma$ is not constant on $[0,t)$. So, $\gamma([0,t))$ is a subset of $A$ containing more than one point, and is connected, contradicting the fact that $A$ is totally disconnected. Hence, $W=\emptyset$.

So, we have constructed a disjoint nonempty pair $U,V$ of open subsets of $X$ such that $C=X\setminus (U\cup V)$ is contained in $A$ and, therefore, is totally disconnected. That, is $C$ is a totally disconnected closed set which disconnects $X$. I'll prove that this is impossible using a bit of simple intersection theory.

Let $S$ be a closed subset of $C$ such that $C\setminus S$ is closed. If $\gamma\colon[0,1]\to X$ is a path with $\gamma(0),\gamma(1)\in X\setminus S$, then we can define the intersection number of $\gamma$ with $S$ as follows. Choose $0=t_0\le t_1\le\cdots\le t_n=1$ such that each $\gamma(t_k)\in X\setminus S$ and $\gamma([t_{k-1},t_k])$ is contained in one of the open sets $X\setminus S$ or $X\setminus (C\setminus S)$. On the interval $[t_{k-1},t_k]$ we can assign an intersection number of $0$ if $\gamma([t_{k-1},t_k])\subseteq X\setminus S$, otherwise we assign the number $F(\gamma(t_k))-F(\gamma(t_{k-1}))$, where $F=1$ on $V$ and $F=0$ on $U$. Sum these up to get the intersection number of $\gamma$ with $S$. It can be seen that adding additional points to the $t_k$ does not change the intersection number, so it is independent of the choice of the $t_k$. It can also be seen that the intersection number will be unchanged under small changes in the path, so homotopic paths have the same intersection numbers.

Now choose an arc $\gamma\colon[0,1]\to X$ with $\gamma(0)\in U$ and $\gamma(1)\in V$, and let $t$ be the supremum of $\gamma^{-1}(U)$. By the hypothesis, there is a path $\tilde \gamma\colon[0,1]\to X\setminus\{\gamma(t)\}$ with $\tilde\gamma(0)=\gamma(0)$ and $\tilde\gamma(1)=\gamma(1)$. As $\gamma^{-1}(C)$ is totally disconnected, there are points $t_0 < t < t_1$ arbitrarily close to $t$ such that $\gamma(t_0)\in U$, $\gamma(t_1)\in V$ and, choosing them close enough, $\gamma([t_0,t_1])$ will be disjoint from the image of $\tilde\gamma$. Then, as $T_1=\gamma([t_0,t_1])\cap C$ and $T_2=(\gamma([0,t_0])\cup\gamma([t_1,1])\cup\tilde\gamma([0,1]))\cap C$ are disjoint compact subsets of $C$, we can use total disconnectedness to find a closed $S\subset C$ containing $T_1$ with $C\setminus S$ closed and containing $T_2$. Then, the intersection number of $\gamma$ with $S$ is 1 and the intersection number of $\tilde\gamma$ with $S$ is $0$, so the paths are not homotopic, contradicting simply connectedness of $X$.

Here's a proof that if $X$ is any simply connected Hausdorff space such that $X\setminus \{p\}$ is path connected for all $p\in X$, then the complement of any totally disconnected subset is connected. In particular, if $X$ has more than one point then it cannot be the disjoint union of two totally disconnected subsets. Taking $X=\mathbb{R}^2$ answers the question.

[This is inspired by Włodzimierz Holsztyński's first answer, by trying to prove the result in a similar way, but using as few advanced properties of $\mathbb{R}^2$ as possible. I expect that the connectedness properties can be replaced by a statement about the Čech cohomology, such as $\check{H}^1(X)=0$ and $X\setminus\{p\}$ is connected for each $p$].

I will use proof by contradiction. So, suppose that $A\subseteq X$ is totally disconnected such that $B\equiv X\setminus A$ is not connected. Then, there are open $U,V\subseteq X$ such that $B\cap U$, $B\cap V$ are disjoint and nonempty and such that $B\subseteq U\cup V$. So, $W= U\cap V$ is disjoint from $B$ and, hence, $W\subseteq A$. If $W$ was nonempty, then choosing points $p\in W$ and $q\in X\setminus\{p\}$, there is a continuous $\gamma\colon[0,1]\to X$ with $\gamma(0)=p$, $\gamma(1)=q$. Letting $t$ be maximal such that $[0,t)\subseteq\gamma^{-1}(W)$ then $t > 0$ and the Hausdorff hypothesis implies that $\gamma$ is not constant on $[0,t)$. So, $\gamma([0,t))$ is a subset of $A$ containing more than one point, and is connected, contradicting the fact that $A$ is totally disconnected. Hence, $W=\emptyset$.

So, we have constructed a disjoint nonempty pair $U,V$ of open subsets of $X$ such that $C=X\setminus (U\cup V)$ is contained in $A$ and, therefore, is totally disconnected. That, is $C$ is a totally disconnected closed set which disconnects $X$. I'll prove that this is impossible using a bit of simple intersection theory.

Let $S$ be a closed subset of $C$ such that $C\setminus S$ is closed. If $\gamma\colon[0,1]\to X$ is a path with $\gamma(0),\gamma(1)\in X\setminus S$, then we can define the intersection number of $\gamma$ with $S$ as follows. Choose $0=t_0\le t_1\le\cdots\le t_n=1$ such that each $\gamma(t_k)\in X\setminus S$ and $\gamma([t_{k-1},t_k])$ is contained in one of the open sets $X\setminus S$ or $X\setminus (C\setminus S)$. On the interval $[t_{k-1},t_k]$ we can assign an intersection number of $0$ if $\gamma([t_{k-1},t_k])\subseteq X\setminus S$, otherwise we assign the number $F(\gamma(t_k))-F(\gamma(t_{k-1}))$, where $F=1$ on $V$ and $F=0$ on $U$. Sum these up to get the intersection number of $\gamma$ with $S$. It can be seen that adding additional points to the $t_k$ does not change the intersection number, so it is independent of the choice of the $t_k$. It can also be seen that the intersection number will be unchanged under small changes in the path, so homotopic paths have the same intersection numbers.

Now choose an arc $\gamma\colon[0,1]\to X$ with $\gamma(0)\in U$ and $\gamma(1)\in V$, and let $t$ be the supremum of $\gamma^{-1}(U)$. By the hypothesis, there is a path $\tilde \gamma\colon[0,1]\to X\setminus\{\gamma(t)\}$ with $\tilde\gamma(0)=\gamma(0)$ and $\tilde\gamma(1)=\gamma(1)$. As $\gamma^{-1}(C)$ is totally disconnected, there are points $t_0 < t < t_1$ arbitrarily close to $t$ such that $\gamma(t_0)\in U$, $\gamma(t_1)\in V$ and, choosing them close enough, $\gamma([t_0,t_1])$ will be disjoint from the image of $\tilde\gamma$. Then, as $T_1=\gamma([t_0,t_1])\cap C$ and $T_2=(\gamma([0,t_0])\cup\gamma([t_1,1])\cup\tilde\gamma([0,1]))\cap C$ are disjoint compact subsets of $C$, we can use total disconnectedness to find a closed $S\subset C$ containing $T_1$ with $C\setminus S$ closed and containing $T_2$. Then, the intersection number of $\gamma$ with $S$ is 1 and the intersection number of $\tilde\gamma$ with $S$ is $0$, so the paths are not homotopic, contradicting simply connectedness of $X$.

Here's a proof that if $X$ is any simply connected Hausdorff space such that $X\setminus \{p\}$ is path connected for all $p\in X$, then the complement of any totally disconnected subset is connected. In particular, if $X$ has more than one point then it cannot be the disjoint union of two totally disconnected subsets. In particular, we can take $X=\mathbb{R}^2$ to answer the question.

[This is inspired by Włodzimierz Holsztyński's first answer, by trying to prove the result in a similar way, but using as few advanced properties of $\mathbb{R}^2$ as possible. I expect that the connectedness properties can be replaced by a statement about the Čech cohomology, such as $\check{H}^2(X)=0$ and $X\setminus\{p\}$ is connected for each $p$].

I will use proof by contradiction. So, suppose that $A\subseteq X$ is totally disconnected such that $B\equiv X\setminus A$ is not connected. Then, there are open $U,V\subseteq X$ such that $B\cap U$, $B\cap V$ are disjoint and nonempty and such that $B\subseteq U\cup V$. So, $W= U\cap V$ is disjoint from $B$ and, hence, $W\subseteq A$. If $W$ was nonempty, then choosing points $p\in W$ and $q\in X\setminus\{p\}$, there is a continuous $\gamma\colon[0,1]\to X$ with $\gamma(0)=p$, $\gamma(1)=q$. Letting $t$ be maximal such that $[0,t)\subseteq\gamma^{-1}(W)$ then $t > 0$ and the Hausdorff hypothesis implies that $\gamma$ is not constant on $[0,t)$. So, $\gamma([0,t))$ is a subset of $A$ containing more than one point, and is connected, contradicting the fact that $A$ is totally disconnected. Hence, $W=\emptyset$.

So, we have constructed a disjoint nonempty pair $U,V$ of open subsets of $X$ such that $C=X\setminus (U\cup V)$ is contained in $A$ and, therefore, is totally disconnected. That, is $C$ is a totally disconnected closed set which disconnects $X$. I'll prove that this is impossible using a bit of simple intersection theory.

Let $S$ be a closed subset of $C$ such that $C\setminus S$ is closed. If $\gamma\colon[0,1]\to X$ is a path with $\gamma(0),\gamma(1)\in X\setminus S$, then we can define the intersection number of $\gamma$ with $S$ as follows. Choose $0=t_0\le t_1\le\cdots\le t_n=1$ such that each $\gamma(t_k)\in X\setminus S$ and $\gamma([t_{k-1},t_k])$ is contained in one of the open sets $X\setminus S$ or $X\setminus (C\setminus S)$. On the interval $[t_{k-1},t_k]$ we can assign an intersection number of $0$ if $\gamma([t_{k-1},t_k])\subseteq X\setminus S$, otherwise we assign the number $F(\gamma(t_k))-F(\gamma(t_{k-1}))$, where $F=1$ on $V$ and $F=0$ on $U$. Sum these up to get the intersection number of $\gamma$ with $S$. It can be seen that adding additional points to the $t_k$ does not change the intersection number, so it is independent of the choice of the $t_k$. It can also be seen that the intersection number will be unchanged under small changes in the path, so homotopic paths have the same intersection numbers.

Now choose an arc $\gamma\colon[0,1]\to X$ with $\gamma(0)\in U$ and $\gamma(1)\in V$, and let $t$ be the supremum of $\gamma^{-1}(U)$. By the hypothesis, there is a path $\tilde \gamma\colon[0,1]\to X\setminus\{\gamma(t)\}$ with $\tilde\gamma(0)=\gamma(0)$ and $\tilde\gamma(1)=\gamma(1)$. As $\gamma^{-1}(C)$ is totally disconnected, there are points $t_0 < t < t_1$ arbitrarily close to $t$ such that $\gamma(t_0)\in U$, $\gamma(t_1)\in V$ and, choosing them close enough, $\gamma([t_0,t_1])$ will be disjoint from the image of $\tilde\gamma$. Then, as $T_1=\gamma([t_0,t_1])\cap C$ and $T_2=(\gamma([0,t_0])\cup\gamma([t_1,1])\cup\tilde\gamma([0,1]))\cap C$ are disjoint compact subsets of $C$, we can use total disconnectedness to find a closed $S\subset C$ containing $T_1$ with $C\setminus S$ closed and containing $T_2$. Then, the intersection number of $\gamma$ with $S$ is 1 and the intersection number of $\tilde\gamma$ with $S$ is $0$, so the paths are not homotopic, contradicting simply connectedness of $X$.

Here's a proof that if $X$ is any simply connected Hausdorff space such that $X\setminus \{p\}$ is path connected for all $p\in X$, then the complement of any totally disconnected subset is connected. In particular, if $X$ has more than one point then it cannot be the disjoint union of two totally disconnected subsets. Taking $X=\mathbb{R}^2$ answers the question.

[This is inspired by Włodzimierz Holsztyński's first answer, by trying to prove the result in a similar way, but using as few advanced properties of $\mathbb{R}^2$ as possible. I expect that the connectedness properties can be replaced by a statement about the Čech cohomology, such as $\check{H}^2(X)=0$ and $X\setminus\{p\}$ is connected for each $p$].

I will use proof by contradiction. So, suppose that $A\subseteq X$ is totally disconnected such that $B\equiv X\setminus A$ is not connected. Then, there are open $U,V\subseteq X$ such that $B\cap U$, $B\cap V$ are disjoint and nonempty and such that $B\subseteq U\cup V$. So, $W= U\cap V$ is disjoint from $B$ and, hence, $W\subseteq A$. If $W$ was nonempty, then choosing points $p\in W$ and $q\in X\setminus\{p\}$, there is a continuous $\gamma\colon[0,1]\to X$ with $\gamma(0)=p$, $\gamma(1)=q$. Letting $t$ be maximal such that $[0,t)\subseteq\gamma^{-1}(W)$ then $t > 0$ and the Hausdorff hypothesis implies that $\gamma$ is not constant on $[0,t)$. So, $\gamma([0,t))$ is a subset of $A$ containing more than one point, and is connected, contradicting the fact that $A$ is totally disconnected. Hence, $W=\emptyset$.

So, we have constructed a disjoint nonempty pair $U,V$ of open subsets of $X$ such that $C=X\setminus (U\cup V)$ is contained in $A$ and, therefore, is totally disconnected. That, is $C$ is a totally disconnected closed set which disconnects $X$. I'll prove that this is impossible using a bit of simple intersection theory.

Let $S$ be a closed subset of $C$ such that $C\setminus S$ is closed. If $\gamma\colon[0,1]\to X$ is a path with $\gamma(0),\gamma(1)\in X\setminus S$, then we can define the intersection number of $\gamma$ with $S$ as follows. Choose $0=t_0\le t_1\le\cdots\le t_n=1$ such that each $\gamma(t_k)\in X\setminus S$ and $\gamma([t_{k-1},t_k])$ is contained in one of the open sets $X\setminus S$ or $X\setminus (C\setminus S)$. On the interval $[t_{k-1},t_k]$ we can assign an intersection number of $0$ if $\gamma([t_{k-1},t_k])\subseteq X\setminus S$, otherwise we assign the number $F(\gamma(t_k))-F(\gamma(t_{k-1}))$, where $F=1$ on $V$ and $F=0$ on $U$. Sum these up to get the intersection number of $\gamma$ with $S$. It can be seen that adding additional points to the $t_k$ does not change the intersection number, so it is independent of the choice of the $t_k$. It can also be seen that the intersection number will be unchanged under small changes in the path, so homotopic paths have the same intersection numbers.

Now choose an arc $\gamma\colon[0,1]\to X$ with $\gamma(0)\in U$ and $\gamma(1)\in V$, and let $t$ be the supremum of $\gamma^{-1}(U)$. By the hypothesis, there is a path $\tilde \gamma\colon[0,1]\to X\setminus\{\gamma(t)\}$ with $\tilde\gamma(0)=\gamma(0)$ and $\tilde\gamma(1)=\gamma(1)$. As $\gamma^{-1}(C)$ is totally disconnected, there are points $t_0 < t < t_1$ arbitrarily close to $t$ such that $\gamma(t_0)\in U$, $\gamma(t_1)\in V$ and, choosing them close enough, $\gamma([t_0,t_1])$ will be disjoint from the image of $\tilde\gamma$. Then, as $T_1=\gamma([t_0,t_1])\cap C$ and $T_2=(\gamma([0,t_0])\cup\gamma([t_1,1])\cup\tilde\gamma([0,1]))\cap C$ are disjoint compact subsets of $C$, we can use total disconnectedness to find a closed $S\subset C$ containing $T_1$ with $C\setminus S$ closed and containing $T_2$. Then, the intersection number of $\gamma$ with $S$ is 1 and the intersection number of $\tilde\gamma$ with $S$ is $0$, so the paths are not homotopic, contradicting simply connectedness of $X$.

Here's a proof that if $X$ is any simply connected Hausdorff space such that $X\setminus \{p\}$ is path connected for all $p\in X$, then the complement of any totally disconnected subset is connected. In particular, if $X$ has more than one point then it cannot be the disjoint union of two totally disconnected subsets. In particular, we can take $X=\mathbb{R}^2$ to answer the question.

[This is inspired by Włodzimierz Holsztyński's first answer, by trying to prove the result in a similar way, but using as few advanced properties of $\mathbb{R}^2$ as possible. I expect that the connectedness properties can be replaced by a statement about the Čech cohomology, such as $\check{H}^2(X)=0$ and $X\setminus\{p\}$ is connected for each $p$].

I will use proof by contradiction and, wlog, assume that $X$ has more than one point. So, suppose that $A\subseteq X$ is totally disconnected such that $B\equiv X\setminus A$ is not connected. Then, there are open $U,V\subseteq X$ such that $B\cap U$, $B\cap V$ are disjoint and nonempty and such that $B\subseteq U\cup V$. So, $W= U\cap V$ is disjoint from $B$ and, hence, $W\subseteq A$. If $W$ was nonempty, then choosing points $p\in W$ and $q\in X\setminus\{p\}$, there is a continuous $\gamma\colon[0,1]\to X$ with $\gamma(0)=p$, $\gamma(1)=q$. Letting $t$ be maximal such that $[0,t)\subseteq\gamma^{-1}(W)$ then $t > 0$ and the Hausdorff hypothesis implies that $\gamma$ is not constant on $[0,t)$. So, $\gamma([0,t))$ is a subset of $A$ containing more than one point, and is connected, contradicting the fact that $A$ is totally disconnected. Hence, $W=\emptyset$.

So, we have constructed a disjoint nonempty pair $U,V$ of open subsets of $X$ such that $C=X\setminus (U\cup V)$ is contained in $A$ and, therefore, is totally disconnected. That, is $C$ is a totally disconnected closed set which disconnects $X$. I'll prove that this is impossible using a bit of simple intersection theory.

Let $S$ be a closed subset of $C$ such that $C\setminus S$ is closed. If $\gamma\colon[0,1]\to X$ is a path with $\gamma(0),\gamma(1)\in X\setminus S$, then we can define the intersection number of $\gamma$ with $S$ as follows. Choose $0=t_0\le t_1\le\cdots\le t_n=1$ such that each $\gamma(t_k)\in X\setminus S$ and $\gamma([t_{k-1},t_k])$ is contained in one of the open sets $X\setminus S$ or $X\setminus (C\setminus S)$. On the interval $[t_{k-1},t_k]$ we can assign an intersection number of $0$ if $\gamma([t_{k-1},t_k])\subseteq X\setminus S$, otherwise we assign the number $F(\gamma(t_k))-F(\gamma(t_{k-1}))$, where $F=1$ on $V$ and $F=0$ on $U$. Sum these up to get the intersection number of $\gamma$ with $S$. It can be seen that adding additional points to the $t_k$ does not change the intersection number, so it is independent of the choice of the $t_k$. It can also be seen that the intersection number will be unchanged under small changes in the path, so homotopic paths have the same intersection numbers.

Now choose an arc $\gamma\colon[0,1]\to X$ with $\gamma(0)\in U$ and $\gamma(1)\in V$, and let $t$ be the supremum of $\gamma^{-1}(U)$. By the hypothesis, there is a path $\tilde \gamma\colon[0,1]\to X\setminus\{\gamma(t)\}$ with $\tilde\gamma(0)=\gamma(0)$ and $\tilde\gamma(1)=\gamma(1)$. As $\gamma^{-1}(C)$ is totally disconnected, there are points $t_0 < t < t_1$ arbitrarily close to $t$ such that $\gamma(t_0)\in U$, $\gamma(t_1)\in V$ and, choosing them close enough, $\gamma([t_0,t_1])$ will be disjoint from the image of $\tilde\gamma$. Then, as $T_1=\gamma([t_0,t_1])\cap C$ and $T_2=(\gamma([0,t_0])\cup\gamma([t_1,1])\cup\tilde\gamma([0,1]))\cap C$ are disjoint compact subsets of $C$, we can use total disconnectedness to find a closed $S\subset C$ containing $T_1$ with $C\setminus S$ closed and containing $T_2$. Then, the intersection number of $\gamma$ with $S$ is 1 and the intersection number of $\tilde\gamma$ with $S$ is $0$, so the paths are not homotopic, contradicting simply connectedness of $X$.

Here's a proof that if $X$ is any simply connected Hausdorff space such that $X\setminus \{p\}$ is path connected for all $p\in X$, then the complement of any totally disconnected subset is connected. In particular, if $X$ has more than one point then it cannot be the disjoint union of two totally disconnected subsets. In particular, we can take $X=\mathbb{R}^2$ to answer the question.

[This is inspired by Włodzimierz Holsztyński's first answer, by trying to prove the result in a similar way, but using as few advanced properties of $\mathbb{R}^2$ as possible. I expect that the connectedness properties can be replaced by a statement about the Čech cohomology, such as $\check{H}^2(X)=0$ and $X\setminus\{p\}$ is connected for each $p$].

I will use proof by contradiction. So, suppose that $A\subseteq X$ is totally disconnected such that $B\equiv X\setminus A$ is not connected. Then, there are open $U,V\subseteq X$ such that $B\cap U$, $B\cap V$ are disjoint and nonempty and such that $B\subseteq U\cup V$. So, $W= U\cap V$ is disjoint from $B$ and, hence, $W\subseteq A$. If $W$ was nonempty, then choosing points $p\in W$ and $q\in X\setminus\{p\}$, there is a continuous $\gamma\colon[0,1]\to X$ with $\gamma(0)=p$, $\gamma(1)=q$. Letting $t$ be maximal such that $[0,t)\subseteq\gamma^{-1}(W)$ then $t > 0$ and the Hausdorff hypothesis implies that $\gamma$ is not constant on $[0,t)$. So, $\gamma([0,t))$ is a subset of $A$ containing more than one point, and is connected, contradicting the fact that $A$ is totally disconnected. Hence, $W=\emptyset$.

So, we have constructed a disjoint nonempty pair $U,V$ of open subsets of $X$ such that $C=X\setminus (U\cup V)$ is contained in $A$ and, therefore, is totally disconnected. That, is $C$ is a totally disconnected closed set which disconnects $X$. I'll prove that this is impossible using a bit of simple intersection theory.

Let $S$ be a closed subset of $C$ such that $C\setminus S$ is closed. If $\gamma\colon[0,1]\to X$ is a path with $\gamma(0),\gamma(1)\in X\setminus S$, then we can define the intersection number of $\gamma$ with $S$ as follows. Choose $0=t_0\le t_1\le\cdots\le t_n=1$ such that each $\gamma(t_k)\in X\setminus S$ and $\gamma([t_{k-1},t_k])$ is contained in one of the open sets $X\setminus S$ or $X\setminus (C\setminus S)$. On the interval $[t_{k-1},t_k]$ we can assign an intersection number of $0$ if $\gamma([t_{k-1},t_k])\subseteq X\setminus S$, otherwise we assign the number $F(\gamma(t_k))-F(\gamma(t_{k-1}))$, where $F=1$ on $V$ and $F=0$ on $U$. Sum these up to get the intersection number of $\gamma$ with $S$. It can be seen that adding additional points to the $t_k$ does not change the intersection number, so it is independent of the choice of the $t_k$. It can also be seen that the intersection number will be unchanged under small changes in the path, so homotopic paths have the same intersection numbers.

Now choose an arc $\gamma\colon[0,1]\to X$ with $\gamma(0)\in U$ and $\gamma(1)\in V$, and let $t$ be the supremum of $\gamma^{-1}(U)$. By the hypothesis, there is a path $\tilde \gamma\colon[0,1]\to X\setminus\{\gamma(t)\}$ with $\tilde\gamma(0)=\gamma(0)$ and $\tilde\gamma(1)=\gamma(1)$. As $\gamma^{-1}(C)$ is totally disconnected, there are points $t_0 < t < t_1$ arbitrarily close to $t$ such that $\gamma(t_0)\in U$, $\gamma(t_1)\in V$ and, choosing them close enough, $\gamma([t_0,t_1])$ will be disjoint from the image of $\tilde\gamma$. Then, as $T_1=\gamma([t_0,t_1])\cap C$ and $T_2=(\gamma([0,t_0])\cup\gamma([t_1,1])\cup\tilde\gamma([0,1]))\cap C$ are disjoint compact subsets of $C$, we can use total disconnectedness to find a closed $S\subset C$ containing $T_1$ with $C\setminus S$ closed and containing $T_2$. Then, the intersection number of $\gamma$ with $S$ is 1 and the intersection number of $\tilde\gamma$ with $S$ is $0$, so the paths are not homotopic, contradicting simply connectedness of $X$.