THE RED LINE IS $ \Im(z) = x$

[![enter image description here](https://i.stack.imgur.com/Namct.png)](https://i.stack.imgur.com/Namct.png)

Now, my proof is as follows,

Let $z' \in A$, then take $\epsilon = x - \Im(z')>0$

Now let $w \in D\_{\epsilon}(z')$ and suppose $w \notin A$ then

$$\epsilon > |z'-w| \geq|\Im(z)-\Im(w)| = \color{red}{\Im(w) - \Im(z')} \geq x - \Im(z') = \epsilon$$ A contradiction. This seems all good, but when I thought about it, the red bit seems fishy. For example, in my "diagram", $\Im(w) < Im(z')$, Is there a way to fix this way of proof or is it flawed from the outset?

You see in my diagram if I chosen $w$ above 
$z'$ then it would have been ok.

EDIT:

There are some cool answers already, but I need to show that it is true using the definition.

A set $A$ is open if for all $z \in A$ there exists $r > 0$ such that $D\_r(z) \subset A$