Is $|z-4|\geq |z|$ an open or closed region? After solving $|x+i y-4|\geq |x+iy|$ I found that $|z-4|\geq |z|$ if $x\leq 2$. Does this mean that $|z-4|\geq |z|$ is closed? 
 A: Yes. An alternative way to formalize this, is to not that $f\colon z\mapsto |z-4|-|z|$ is continuous (because subtraction and absolute value are continuous) and hence our region $f^{-1}([0,\infty))$ is closed because it is the pre-image of a closed set under continuous function.
A: We have
$$
\lvert z - 4 \rvert \ge \lvert z \rvert \iff \\
\lvert z - 4 \rvert^2 \ge \lvert z \rvert^2 \iff \\
(x-4)^2 + y^2 \ge x^2 + y^2 \iff \\
-8x + 16 \ge 0 \iff \\
x \le 2
$$
so your solution set agrees with mine.
The question is how to characterize a complex set as open or closed.
According to this definition it would not be open, because e.g. for $z=2$ we could not fit in a disc with positive radius around it and stay within the solution set.
Now for the other property:
$S$ would be closed if its complement is open. The complement would be
$$
S^C = \{ z = x + i y \mid (x,y) \in \mathbb{R}^2, x>2 \}
$$
and this one indeed seems open, as one can fit in positive radius discs around every point of $S^C$.
So the set $S$ is closed and not open.
