Showing a set of the complex plane is closed Determine whether the set 
$$A=\{z\in\mathbb{C}\,\colon \Re{(z)}\ge0,\,\Im{(z)}\ge0,\, \vert z\vert\le1\}$$
is closed.

This is true if its complement is open. We have that
$$\overline{A} = \mathbb{C}\setminus\{z\in\mathbb{C}\,\colon \Re{(z)}\ge0,\,\Im{(z)}\ge0,\, \vert z\vert\le1\}$$
So I have split this complement into a union of finitely many complements: 
$$\overline{A} = \bigcup_{i=1}^3 U_i$$
where
$$U_1=\{z\in\mathbb{C}\colon \Im{(z)}<0\}, \,\,U_2=\{z\in\mathbb{C}\colon\Re{(z)}<0\},\,\,U_3 = \{z\in\mathbb{C}\colon \vert z\vert >1\}$$
Now $U_1,U_2$ are clearly open (omitted detail for brevity). It is left to show $U_3$ is open. That is, if we take $z_0\in U_3$, then we must find an $r>0$ such that $D(z_0,r)\subseteq U_3$. Take 
$$0<r<\vert z_0\vert -1\implies \vert z_0\vert-r>1.$$ 
Let $z\in D(z_0, r)$, then $\vert z_0-z\vert<r$. We want to show $\vert z\vert >1$. Now
\begin{align}
r &> \vert z_0-z\vert \\
&\ge \vert z_0\vert - \vert z\vert
\end{align}
$$\implies \vert z\vert > \vert z_0\vert - r>1$$
as required. Since this was proven for any arbitrary $z\in D(z_0,r)$ we have shown that $D(z_0,r)\subseteq U_3$. Since this was shown for an arbitrary $z_0\in U_5$, hence $U_5$ contains all its interior points and is therefore open. Note $\overline{A}$ is the union of three open sets which is therefore open. Thus $A$ is closed.
 A: You needlessly complicate the problem: denoting by $re(z) = \Re z, \ im(z) = \Im z, \ abs(z) = |z|$, notice that $A = B \cap C \cap D$ where $B = re^{-1} ([0, \infty)), \ C = im^{-1} ([0, \infty)), \ D = abs^{-1} ([0,1])$. Since the functions $re, im, abs$ are continuous and the intervals $[0,\infty)$ and $[0,1]$ are closed in $\Bbb R$, their respective preimages $B,C,D$ will also be closed in $\Bbb C$ (the preimage of a closed set through a continuous function is closed). Since the finite intersection of closed sets is closed, $A = B \cap C \cap D$ will also be closed.
A: Since in the comments you asked for a geometrical proof, here we go:
First of all note that the set $A$ you defined is simply a 'quarter pie/unit ball' in the first quadrant of $\mathbb{R}^2$.
If you take an arbitrary point $z$ in the complemenet of $A$ you can define $r=d(A,z)=\inf_{z_A \in A}d(z_a,z)$, i.e. the shortest distance between $z$ and $A$ (note that here the structure of $A$ is crucial!). Now if you take an $\epsilon >0$ such that $ r> \epsilon >0$ you will obtain that the open ball $B_{\epsilon}=\{z'\in \mathbb{ C}\mid d(z',z) < \epsilon \}$ is completely contained in the complement of $A$.
So we conclude that $z$ is an inner point of $A^c$ and since $z$ was arbitrary we obtain that any $z$ in $A^c$ is an inner point and so $A^c$ is open.
This implies that $A$ is closed.
A: Using converging sequences seems the easiest way to prove it. $Re(z), Im(z)$ and the absolute value are all continuous functions and they all preserve the properties of the set $A$
