Let $A$ be a path-connected subset of $\mathbb R^2$ such that the removal of any singleton from $A$ splits $A$ into two open connected components, each of which is path-connected.
Is $A$ necessarily homeomorphic to $\mathbb{R}$?
Let $A$ be a path-connected subset of $\mathbb R^2$ such that the removal of any singleton from $A$ splits $A$ into two open connected components, each of which is path-connected.
Is $A$ necessarily homeomorphic to $\mathbb{R}$?
Ward has given the following characterization of the real line: a connected, locally connected separable metric space in which each point is a cut point, i.e., its removal splits the space into two connected subsets (Proc. London Math. Soc. 1936). This implies a positive answer to your question, assuming the set has more than one point.
Theorem. Suppose $X$ is a connected separable metric space such that $X\setminus \{x\}$ has exactly two (non-empty) open path-connected components for every $x\in X$. Then $X\simeq \mathbb R$.
Proof. We claim that each point $x\in X$ is contained in an arc $A$ with endpoints $a$ and $b$ such that $x\in A\setminus \{a,b\}$ and the interval $A\setminus \{a,b\}$ is open in $X$. Clearly this would imply $X$ is locally connected, and then by Ward's theorem (see earlier answer) we get $X\simeq \mathbb R$.
Let $x\in X$. Let $y\in X\setminus \{x\}$. It is a simple exercise to prove that $X\setminus \{x,y\}$ is the union of three non-empty disjoint open sets $U\sqcup V\sqcup W$ such that $U\cup \{x\}\cup V$ is path-connected. Let $A$ be an arc with endpoints $a\in U$ and $b\in V$. Then we have $x\in A\setminus \{a,b\}$.
We need to two claims. The proofs are by contradiction.
Claim 1: $X$ contains no simple closed curve. Suppose $S\subseteq X$ is a simple closed curve. Then for every point $x\in S$ there are two non-empty open sets $U_x$ and $V_x$ such that $X\setminus \{x\}=U_x\sqcup V_x$ and $S\subseteq U_x\cup\{x\}$. Notice that the sets $V_x$, $x\in S$, are pairwise disjoint. This contradicts the fact that $X$ is separable.
Claim 2: $X$ contains no simple triod. Suppose $T\subseteq X$ is a simple triod which is the union of three arcs $A_1,A_2,A_3$ and has center vertex $a$. Write $X\setminus \{a\}= U\sqcup V$ so that $U$ and $V$ are path-connected. Without loss of generality $A_1\cup A_2\subseteq U\cup \{a\}$. Let $B\subseteq U$ be an arc meeting both $A_1$ and $A_2$. Then $B\cup A_1\cup A_2$ must contain a simple closed curve. This contradicts Claim 1.
By Claims 1 and 2 we have:
Claim 3. The union of any two intersecting arcs in $X$ is again an arc.
Finally we can show $A\setminus \{a,b\}$ is open in $X$. Suppose to the contrary that $c\in A\setminus \{a,b\}$ is the limit of a sequence of points $x_n\in X\setminus A$. By the path-connected property and Claim 3, there is an increasing collection of arcs $A_0\subseteq A_1\subseteq ...$ such that $\{x_n\}\cup A\subseteq A_n$ for every $n<\omega$. Then we get a one-to-one continuous image of the half-line $[0,\infty)$ which begins in $A$ and also limits to $A$ at the infinity end. This forms a circle-like configuration, which is impossible by the arguments in Claim 2 (the "ray limiting to itself" would remain connected upon the deletion of any point, and this leads to an uncountable collection of pairwise disjoint non-empty open sets). $\;\blacksquare$
Note 1: I used path-connectedness of $X$ near the end. This was a hypothesis in the original question, but it actually follows from my assumptions in the theorem statement. It's a simple exercise.
Note 2: Ward's theorem is probably overkill. We already see that $X$ is an increasing union of arcs and contains no ray limiting to itself.