Let $S \subseteq \mathbb C$ be an open connected set. This implies that it the set is polygon connected. That means, given $a,b\in S$, there is a finite sequence of line segments $[a,a_1],[a_1,a_2],...,[a_{n-1},b]$ which are all contained in $S$.
But is it "grid connected?" (I'm inventing this terminology on the spot, so bear with me.) I define a set $E\subseteq \mathbb C$ to be "grid connected" if, given points $a,b\in E$, there is a finite sequence of line segments $[a,a_1],[a_1,a_2],...,[a_{n-1},b]$ each of which is either horizontal or vertical.
For example, the plane $\mathbb C$ is "grid connected" because, given distinct points $a+bi$ and $c+di$, we can connect them with the line segments $[a+bi, c+bi]$ and $[c+bi,c+di]$. (i.e. with a horizontal line segment and a vertical line segment.
Since you agree that we can connect using line segments, it suffices to prove that for each segment $[z,w]$ there are a finite number of vertical or horizontal segments $[a_1,a_2]$, $[a_2,a_3]$, ... $[a_{k-1}, a_k]$, contained in the original set $S$ such that $a_1 = z$ and $a_k = w$.
Since $z\in S$ and $S$ is open, there is an open neighborhood of $z$ contained in $S$. There we can connect with horizontal and vertical lines as shown in the picture, and, importantly, we're still in $S$.
Finally, since $[z,w]$ is compact, only finitely many of those neighborhoods are needed.