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.