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I was looking for a related fact, and surprisingly couldn't find anything relevant, except of this question. Even though it was answered 10 years ago, perhaps the following result could be useful to somebody.

Proposition. Let $X$ be a connected metric space that contains more than one point. Let $A\subset X$ be totally disconnected and locally compact with respect to the subspace topology. Then $A$ is nowhere dense.

Proof. First, let us show $int A =\varnothing$. Assume that $U$ is an open set in $X$ such that $\overline{U}$ is compact and contained in $A$, and $x\in U$. Since a totally disconnected locally compact paracompact space is zero-dimensional (see 6.2.9 in Engelking's General Topology), there is an open neighborhood $V\subset U$ of $x$ that is clopen in $A$. Then, there are closed set $F$ in $X$ and open set $W$ in $X$ such that $V=A\cap F=A\cap W$. Since $V\subset U\subset A$, $V=U\cap W$ is open in $X$. Since $V\subset \overline{U}\subset A$, $V=\overline{U}\cap F$ is closed in $X$. Hence, $V$ is nonempty and clopen in $X$ which contradicts its connectedness.

Now recall that a locally compact set is open in its closure (see 3.3.9 in Engelking). Hence, $A=\overline{A}\cap U$, for some open $U\subset X$. Assume $int \overline{A}\ne\varnothing$. Then, there is $x\in int \overline{A} \cap A= int \overline{A} \cap\overline{A}\cap U=int \overline{A} \cap U\subset \overline{A} \cap U=A$, from where $x\in int \overline{A} \cap U\subset A$, and so $x\in int A$. Contradiction with the previous step.

Corollary. If $X$ is a connected complete metric space that contains more than one point, it cannot be covered by a countable collection of totally disconnected locally compact subsets.

Also, the answer to the original question is negative since a point is totally disconnected.

I was looking for a related fact, and surprisingly couldn't find anything relevant, except of this question. Even though it was answered 10 years ago, perhaps the following result could be useful to somebody.

Proposition. Let $X$ be a connected metric space that contains more than one point. Let $A\subset X$ be totally disconnected and locally compact with respect to the subspace topology. Then $A$ is nowhere dense.

Proof. First, let us show $int A =\varnothing$. Assume that $U$ is an open set in $X$ such that $\overline{U}$ is compact and contained in $A$, and $x\in U$. Since a totally disconnected locally compact paracompact space is zero-dimensional (see 6.2.9 in Engelking's General Topology), there is an open neighborhood $V\subset U$ of $x$ that is clopen in $A$. Then, there are closed set $F$ in $X$ and open set $W$ in $X$ such that $V=A\cap F=A\cap W$. Since $V\subset U\subset A$, $V=U\cap W$ is open in $X$. Since $V\subset \overline{U}\subset A$, $V=\overline{U}\cap F$ is closed in $X$. Hence, $V$ is nonempty and clopen in $X$ which contradicts its connectedness.

Now recall that a locally compact set is open in its closure (see 3.3.9 in Engelking). Hence, $A=\overline{A}\cap U$, for some open $U\subset X$. Assume $int \overline{A}\ne\varnothing$. Then, there is $x\in int \overline{A} \cap A= int \overline{A} \cap\overline{A}\cap U=int \overline{A} \cap U\subset \overline{A} \cap U=A$, from where $x\in int \overline{A} \cap U\subset A$, and so $x\in int A$. Contradiction with the previous step.

Corollary. If $X$ is a connected complete metric space that contains more than one point, it cannot be covered by a countable collection of totally disconnected locally compact subsets.

In particular, we cannot remove a closed totally disconnected set (e.g. a single point) from a continuum to make it totally disconnected.

I was looking for a related fact, and surprisingly couldn't find anything relevant, except of this question. Even though it was answered 10 years ago, perhaps the following result could be useful to somebody.

Proposition. Let $X$ be a connected metric space. Let $A\subset X$ be totally disconnected and locally compact with respect to the subspace topology. Then $A$ is nowhere dense.

Proof. First, let us show $int A =\varnothing$. Assume that $U$ is an open set in $X$ such that $\overline{U}$ is compact and contained in $A$, and $x\in U$. Since a totally disconnected locally compact paracompact space is zero-dimensional (see 6.2.9 in Engelking's General Topology), there is an open neighborhood $V\subset U$ of $x$ that is clopen in $A$. Then, there are closed set $F$ in $X$ and open set $W$ in $X$ such that $V=A\cap F=A\cap W$. Since $V\subset U\subset A$, $V=U\cap W$ is open in $X$. Since $V\subset \overline{U}\subset A$, $V=\overline{U}\cap F$ is closed in $X$. Hence, $V$ is nonempty and clopen in $X$ which contradicts its connectedness.

Now recall that a locally compact set is open in its closure (see 3.3.9 in Engelking). Hence, $A=\overline{A}\cap U$, for some open $U\subset X$. Assume $int \overline{A}\ne\varnothing$. Then, there is $x\in int \overline{A} \cap A= int \overline{A} \cap\overline{A}\cap U=int \overline{A} \cap U\subset \overline{A} \cap U=A$, from where $x\in int \overline{A} \cap U\subset A$, and so $x\in int A$. Contradiction with the previous step.

Corollary. If $X$ is a complete metric space, it cannot be covered by a countable collection of totally disconnected locally compact subsets.

Also, the answer to the original question is negative since a point is totally disconnected.

I was looking for a related fact, and surprisingly couldn't find anything relevant, except of this question. Even though it was answered 10 years ago, perhaps the following result could be useful to somebody.

Proposition. Let $X$ be a connected metric space that contains more than one point. Let $A\subset X$ be totally disconnected and locally compact with respect to the subspace topology. Then $A$ is nowhere dense.

Proof. First, let us show $int A =\varnothing$. Assume that $U$ is an open set in $X$ such that $\overline{U}$ is compact and contained in $A$, and $x\in U$. Since a totally disconnected locally compact paracompact space is zero-dimensional (see 6.2.9 in Engelking's General Topology), there is an open neighborhood $V\subset U$ of $x$ that is clopen in $A$. Then, there are closed set $F$ in $X$ and open set $W$ in $X$ such that $V=A\cap F=A\cap W$. Since $V\subset U\subset A$, $V=U\cap W$ is open in $X$. Since $V\subset \overline{U}\subset A$, $V=\overline{U}\cap F$ is closed in $X$. Hence, $V$ is nonempty and clopen in $X$ which contradicts its connectedness.

Now recall that a locally compact set is open in its closure (see 3.3.9 in Engelking). Hence, $A=\overline{A}\cap U$, for some open $U\subset X$. Assume $int \overline{A}\ne\varnothing$. Then, there is $x\in int \overline{A} \cap A= int \overline{A} \cap\overline{A}\cap U=int \overline{A} \cap U\subset \overline{A} \cap U=A$, from where $x\in int \overline{A} \cap U\subset A$, and so $x\in int A$. Contradiction with the previous step.

Corollary. If $X$ is a connected complete metric space that contains more than one point, it cannot be covered by a countable collection of totally disconnected locally compact subsets.

Also, the answer to the original question is negative since a point is totally disconnected.

I was looking for a related fact, and surprisingly couldn't find anything relevant, except of this question. Even though it was answered 10 years ago, perhaps the following result could be useful to somebody.

Proposition. Let $X$ be a connected metric space. Let $A\subset X$ be totally disconnected and locally compact with respect to the subspace topology. Then $int A =\varnothing$.

Proof. Assume that $U$ is an open set in $X$ such that $\overline{U}$ is compact and contained in $A$, and $x\in U$. Since a totally disconnected locally compact paracompact space is zero-dimensional (see 6.2.9 in Engelking's General Topology), there is an open neighborhood $V\subset U$ of $x$ that is clopen in $A$. Then, there are closed set $F$ in $X$ and open set $W$ in $X$ such that $V=A\cap F=A\cap W$. Since $V\subset U\subset A$, $V=U\cap W$ is open in $X$. Since $V\subset \overline{U}\subset A$, $V=\overline{U}\cap F$ is closed in $X$. Hence, $V$ is nonempty and clopen in $X$ which contradicts its connectedness.

Corollary. If $X$ is a complete metric space, it cannot be covered by a countable collection of closed totally disconnected locally compact subsets.

Also, the answer to the original question is negative since adding a point cannot fix the empty interior in a metric space.

I was looking for a related fact, and surprisingly couldn't find anything relevant, except of this question. Even though it was answered 10 years ago, perhaps the following result could be useful to somebody.

Proposition. Let $X$ be a connected metric space. Let $A\subset X$ be totally disconnected and locally compact with respect to the subspace topology. Then $A$ is nowhere dense.

Proof. First, let us show $int A =\varnothing$. Assume that $U$ is an open set in $X$ such that $\overline{U}$ is compact and contained in $A$, and $x\in U$. Since a totally disconnected locally compact paracompact space is zero-dimensional (see 6.2.9 in Engelking's General Topology), there is an open neighborhood $V\subset U$ of $x$ that is clopen in $A$. Then, there are closed set $F$ in $X$ and open set $W$ in $X$ such that $V=A\cap F=A\cap W$. Since $V\subset U\subset A$, $V=U\cap W$ is open in $X$. Since $V\subset \overline{U}\subset A$, $V=\overline{U}\cap F$ is closed in $X$. Hence, $V$ is nonempty and clopen in $X$ which contradicts its connectedness.

Now recall that a locally compact set is open in its closure (see 3.3.9 in Engelking). Hence, $A=\overline{A}\cap U$, for some open $U\subset X$. Assume $int \overline{A}\ne\varnothing$. Then, there is $x\in int \overline{A} \cap A= int \overline{A} \cap\overline{A}\cap U=int \overline{A} \cap U\subset \overline{A} \cap U=A$, from where $x\in int \overline{A} \cap U\subset A$, and so $x\in int A$. Contradiction with the previous step.

Corollary. If $X$ is a complete metric space, it cannot be covered by a countable collection of totally disconnected locally compact subsets.

Also, the answer to the original question is negative since a point is totally disconnected.