I think this works but haven't checked. I'm pretty sure that for any open set it's easy to find a map that takes any open subset of that set to all of $\mathbb{C}$, or to all of a square, or to whatever single set you feel like. So now for each n choose a map that takes every open subset of the annulus {$z: n < |z| \leq n+1$} to the square that consists of all points with real and imaginary parts less than or equal to n. Now, given any triangle, there will be a maximum n such that it belongs to the nth annulus, and it will intersect that annulus in an open set and therefore map to a square.

I think this works but haven't checked. I'm pretty sure that for any open set it's easy to find a map that takes any open subset of that set to all of $\mathbb{C}$, or to all of a square, or to whatever single set you feel like. So now for each n choose a map that takes every open subset of the annulus $\{z: n < |z| \leq n+1\}$ to the square that consists of all points with real and imaginary parts less than or equal to n. Now, given any triangle, there will be a maximum n such that it belongs to the nth annulus, and it will intersect that annulus in an open set and therefore map to a square.