Show that a finite intersection of open sets in $\mathbb{C}$ is an open set in $\mathbb{C}$.
Attempt: I want to show $\bigcap_{i=0}^{n}A_i$ is open. Let $z\in\bigcap^{n}A_i$ for open $A_i$ in $\mathbb{C}$. Then, for some $i$, $z\in A_n$. Since $A_i$ is open, $\exists\epsilon>0$ s.t. $B_{\epsilon}(z)\subseteq A_i$. But that means $B_{\epsilon}(z)\subseteq A_i\subseteq \bigcap^{n}A_i$ thus $\bigcap^{n}A_i$ is open.
A finite union of closed sets in $\mathbb{C}$ is a closed set in $\mathbb{C}$.
Attempt: I want to show that $\bigcup_{i=0}^{n}A_i$ for closed $A_i$ is closed, so I'll show ($\bigcup_{i=0}^{n}A_i )^{c}$ is open. Let $z\in(\bigcup_{i=0}^{n}A_i )^{c}$. Then $z\in A_{i}^{c}$ for some $i$. Since all $A_{i}$ are closed, $A_{i}^{c}$ is open. Then I'm guessing you show that since you can find an open ball around $z$ in $A_{i}^{c}$, you can find an open ball around the point in ($\bigcup_{i=0}^{n}A_i )^{c}$, thus the complement is open?