Prove that $V=\{y \in U |$ exists a path to $x_0$ in U $\}$ is open.

Question

Let an open set $U \subset \mathbb{R^2}$. Let $x_0 \in U$. Prove that $V=\{y \in U |$ exists a path to $x_0$ in U $\}$ is open.

My attempt Let $z \in V$, I need to find a $\epsilon >0$ such that $B_{\epsilon}(z) \subset V$. Since $z \in V$ then $ z\in U$ and $\exists$ $\sigma:[0,1] \rightarrow U $ continuous such that $\sigma(0)=z$, $\sigma(1)=x_0$. Since $z \in U$ exists $\epsilon > 0$ such that $B_{\epsilon}(z) \subset U$. So what I understand is $B_{\epsilon}(z) \subset V$ but I don't understand how use $\sigma$. Could you guide me? please.

Answer 1

Since $U$ is open, there exists $\epsilon > 0$ such that $B_\epsilon(z) \subseteq U$. We can show that $B_\epsilon(z) \subseteq V$ as well.

I will sketch the proof informally. Given $w \in B_\epsilon(z)$, consider the straight line path from $w$ to $z$, call it $\gamma$. Then there is a path from $w$ to $x_0$, which is the joining of the paths $\gamma$ and $\sigma$: we first go from $w$ to $z$ via $\gamma$, then we go from $z$ to $x_0$ via $\sigma$. Thus, $w \in V$ as well.

Can you fill in the details?