Connected Open Subset of Euclidean Space is Path-Connected

Theorem
Let $\R^n$ be a Euclidean $n$-space.

Let $U$ be a connected open subset of $\R^n$.

Then $U$ is path-connected.

Proof
Let $a \in U$.

Let $H \subseteq U$ be the subset of points in $U$ which can be joined to $a$ by a path in $U$.

Let $K = U - H$.

Let $x \in H$.

Then $\exists \epsilon > 0: N_\epsilon \left({x}\right) \subseteq U$, where $N_\epsilon \left({x}\right)$ is the $\epsilon$-neighborhood of $x$.

Given any $y \in N_\epsilon \left({x}\right)$, there is a (straight line) path $g$ in $N_\epsilon \left({x}\right) \subseteq U$ connecting $x$ and $y$.

But since $x \in H$, there is a path $f$ in $U$ joining $a$ to $x$.

From Joining Paths makes Another Path, the path formed by traversing $f$ then $g$ is a path from $a$ to $y$.

So $y \in H$, so $N_\epsilon \left({x}\right) \subseteq H$ and so $H$ is open.

By a similar argument, $K$ is also shown to be open:

If $x \in K$, then $N_\epsilon \left({x}\right) \subseteq U$ for some $\epsilon > 0$.

If any point in $N_\epsilon \left({x}\right)$ can be joined to $a$ by a path in $U$, then so could $x$.

It is clear that $H \cap K = \varnothing$ and $H \cup K = U$ by definition of set difference.

By the fact that $U$ is connected, it follows that $K = \varnothing$, and $H = U$.

Hence the result.