Let us choose some point $x$ of an arbitrary connected open set $U$. Take the subset $A$ of $U$ to be the set of all points $a$ such that $x$ and $a$ are path-connected. We realize that $A$ is simply the path-component of $U$ containing $x$.

By Theorem 25.4, for each open set of $X$, each path-component is open in $X$. Thus, $A$ is open in $X$, and thus, open in $U$. By Theorem 25.5, the components and path-components of $X$ are the same. Since the components of a set will be closed, it follows that the path-components of $X$ are also closed. Thus, $A$ is both open and closed in $U$.

Since $U$ is connected, the only subset of $U$ that is both open and closed in $U$ is $U$ itself. Thus, $A \ = \ U$. This implies that each pair of points in $U$ connected by a path, making $U$ path-connected.