Let us call the set of all $y$, such that there exists a path between $x_0$ and $y$, $A$. Assume that $A$ isn't closed. Consider the set $\bar{A}$. For some $a \ \in \ \bar{A}$, it holds true that each neighbourhood of $a$ intersects $A$. Since $A$ isn't closed, it follows that $A \ \neq \ \bar{A}$. Thus, there is some $b \ \in \ \bar{A}$ such that $b \ \notin A$. Note that $b \ \in \ U$, as $U \ = \ \bar{U}$, since $U$ is connected. Consider an aribtrary open ball around $b$ in $B_d(b, \ \epsilon)$. This open ball must intersect $A$ at at least one point, $c$. Since $c$ is in $A$, it follows by definition that there is a path from $x_0$ to $c$. We also know that open balls are path connected, so there is a path from $c$ to $b$. The map formed from pasting these two paths together, thus going from $x_0$ to $b$, is clearly continuous. Thus, there is a path from $x_0$ to $b$, and $b$ isn't in $A$. This is a contradiction, so it follows that $A \ = \ \bar{A}$ and $A$ is closed.

We now attempt to prove that $A$ is open. We pick some arbitrary point $a$ of $A$. Since $U$ is open, we know that there exists some open ball around $a$, which we call $B_d(a, \ \delta)$, such that $B_d(a, \ \delta) \ \subset \ U$. Let us pick some point $b \ \in \ B_d$. We know that there exists a path between $a$ and $b$, as open balls are path connected. We also know that there exists a path from $x_0$ to $a$. It follows that $x_0$ and $b$ are connected by a path, and $B_d \ \subset \ A$. Thus, for each point $a$ of $A$, there exists an open set of $U$, such that $a \ \in \ B_d \ \subset \ A$. By definition, $A$ is open in $U$.

Since $A$ is both open and closed in $U$, and in a connected set, the only set that is both open and closed is the full set itself, it follows that $A \ = \ U$, and $U$ is path-connected.