Trying to prove the sequence is cauchy If $\Omega_1 \supset \Omega_2 \supset ... \supset \Omega_n \supset ... $ is a sequence of non-empty compact sets in $\mathbb{C}$ with that property that
$$ \lim_{n \to \infty} diam( \Omega_n ) = 0 $$
where $diam A = \sup_{a,b \in A } |a-b| $.
Choose a point $z_n $ in each $\Omega_n$.
Goal: Show that $(z_n)$ is a Cauchy sequence:
Attempt:
Let $\epsilon > 0$ be given. We can find some $N > 0$ such that for all $n > N$, $diam ( \Omega_n ) < \epsilon $. In other words,
$$ \sup_{z_n, w \in \Omega_n} |z_n - w | < \epsilon $$
so, there is some $w' \in \Omega_n $ such that $|z_n - w'| < \epsilon $. Hence, $z_n \to w'$ so $(z_n)$ is Cauchy.
Is this correct? I feel this proof is not fully correct. Any help would be greatly appreciated.

Note since it's a decreasing family and by your choice of $z_n$, $z_n\in \Omega_N\, \forall n\ge N$. Hence if you have $N$, such that $diam ( \Omega_N ) < \epsilon $, that is, 
$$ \sup_{z, w \in \Omega_N} |z - w | < \epsilon $$
Then $|z_n-z_m|< \epsilon\, \forall n,m\ge N$, which means it's Cauchy.
