Let $a \in \mathbb{C}$ and $r>0$. Prove that $\overline{B(z,r)}=B[z,r]$.
I know it's trivial, but I can not find any way to prove it. I started with the definition of "Open Ball", and I did not advance further. Is there any way to prove this? What would be the right way to do it?.
First, note that $B[z, r]$ is a closed set (I'm guessing you've learned this?) that contains $B(z, r)$. Therefore, $\overline{B(z, r)} \subseteq B[z, r]$ (the reason for this varies depending on how you define closures).
On the other hand, if $w \in B[z, r]$, then either $w \in B(z, r)$, in which case we're done, or $w \in S[z, r]$. In this case, note that the sequence $$w_n = z + \left(1 - \frac{1}{n}\right)(w - z)$$ lies in $B(z, r)$, since $$|w_n - z| = \left|\left(1 - \frac{1}{n}\right)(w - z)\right| = \left(1 - \frac{1}{n}\right)r < r,$$ but $w_n \to w$, since $$|w_n - w| = \frac{|z-w|}{n} \to 0.$$ Therefore, $w \in \overline{B(z, r)}$, and we're done.