Let $X$ be a non-empty locally path connected topological space and $A$ a non-empty open subspace of $X$. Is $A$ also locally path connected? I remember seeing this in a lecture but it was used implicitly and without justification. I cannot find a counterexample. So I'm trying to prove it.
Consider any point $x\in A$. To show that $A$ is locally path connected, it suffices to find a neighborhood base at $x$ consisting of path connected subsets of $A$. Since $X$ is locally path connected, we can find a neighborhood base $\mathscr{B}$ at $x$ which consists of path connected subsets of $X$. Let $\mathscr{B}'=\{B\in \mathscr{B}:B\subseteq A\}$. Then $\mathscr{B}'$ is a neighborhood base at $x$ in $A$. Indeed, since $A$ is open in $X$, it is a neighborhood of $x$ in $X$, hence contains some member $B_1$ of $\mathscr{B}$, meaning that $\mathscr{B}'\neq \emptyset$.
Then consider any neighborhood $U$ of $x$ in $A$. We know that $U=A\cap V$ for some neighborhood $V$ of $x$ in $X$. There exists $B_2\in \mathscr{B}$ with $B_2\subseteq V$. By definition of neighborhood base, there exists $B_3\in \mathscr{B}$ with $B_3\subseteq B_1\cap B_2$. It follows that $B_3\subseteq A\cap V=U$. Hence every neighborhood of $x$ in $A$ contains some member of $\mathscr{B}'$, proving the claim.
Now consider any $B\in \mathscr{B}'$ and any two points $y,z\in B$. Since $B$ is a path connected subset of $X$, there exists a continuous function $f:[0,1]\to B$ with $f(0)=y,f(1)=z$. But $f([0,1])\subseteq B\subseteq A$, meaning that $f$ is also a $yz$-path in $A$. We conclude that $A$ is locally path connected.
Is my proof correct? Thanks for any comment.
