If $E_{1},E_{2},\ldots,E_{n}$ are a finite collection of open sets in $X$, then $E_{1}\cap E_{2}\cap\ldots\cap E_{n}$ is also open. If $F_{1},F_{2},\ldots,F_{n}$ is a finite collection of closed sets in $X$, then $F_{1}\cup F_{2}\cup\ldots\cup F_{n}$ is also closed.
MY ATTEMPT
First approach
Let us consider that $x\in E = E_{1}\cap E_{2}\cap\ldots\cap E_{n}$. Then $x\in E_{k}$ for each $1\leq k \leq n$. Since each set $E_{k}$ is open, there are open balls such that $x\in B(x,r_{k})\subseteq E_{k}$.
Hence, if we take $r = \min\{r_{1},r_{2},\ldots,r_{n}\}$, the open ball $B(x,r)$ is contained in $E$. We have just proven that each point of $E$ is an interior point of $E$. Thus $E$ is open.
On the other hand, since each $F_{k}$ is closed, we can rewrite them as $F_{k} = X\cap A^{c}_{k}$, where each $A_{k}$ is open.
Consequently, the set $$ F = (X\cap A^{c}_{1})\cup(X\cap A^{c}_{2})\cup\ldots\cup(X\cap A^{c}_{n}) = X\cap(A^{c}_{1}\cup A^{c}_{2}\cup\ldots\cup A^{c}_{n}) = X\cap A^{c}$$
is closed, where $A = A_{1}\cap A_{2}\cap\ldots\cap A_{n}$ is an open set according to the first part of the proposition.
Second approach
We shall prove first that $\overline{X\cup Y} = \overline{X}\cup\overline{Y}$ using the concept of adherent point.
If $x\in\overline{X\cup Y}$, then for every $r > 0$, there exists an open ball such that $B(x,r)\cap(X\cup Y)\neq\varnothing$.
If $x\in X$, then $x\in\overline{X}$. Otherwise, $x\in\overline{Y}$. Thus we have proven that $x\in\overline{X}\cup\overline{Y}$.
Suppose conversely that $x\in\overline{X}\cup\overline{Y}$.
If $x\in\overline{X}$, there exists an open ball such that $B(x,r)\cap (X\cup Y)\supseteq B(x,r)\cap X \neq \varnothing$ for every $r > 0$.
Thus $x\in\overline{X\cup Y}$.
Similarly, if $x\in\overline{Y}$, there exists an open ball such that $B(x,r)\cap(X\cup Y)\supseteq B(x,r)\cap Y\neq \varnothing$ for every $r > 0$.
Hence $x\in\overline{X\cup Y}$.
We have just proven that $\overline{X}\cup\overline{Y}\subseteq\overline{X\cup Y}$. Therefore $\overline{X}\cup\overline{Y} = \overline{X\cup Y}$.
It is possible then to prove through induction that $X_{1}\cup X_{2}\cup\ldots\cup X_{n}$ is closed whenever $X_{k}$ is closed.
Finally, one has that \begin{align*} F_{1}\cup F_{2}\cup\ldots F_{n} = \overline{F}_{1}\cup\overline{F}_{2}\cup\ldots\cup\overline{F}_{n} = \overline{F_{1}\cup F_{2}\cup\ldots\cup F_{n}} \end{align*} thence we conclude the union is also closed.
The first argument for open sets (using metric balls, and the minimal radius argument) is fine and standard. The closed case follows by de Morgan, using the duality between open and closed sets.
In the second approach, the inclusion $\overline{X \cup Y} \subseteq \overline{X} \cup \overline{Y}$ is shown the wrong way: you rightly say that for every $r>0$ we have $B(x,r) \cap (X \cup Y) \neq \emptyset$, but then distinguish $x \in X$ or $x \in Y$ as the only cases, which is not enough.
A better approach: suppose $x \notin \overline{X} \cup \overline{Y}$, then there is an $r_1 > 0$ such that $B(x,r_1) \cap X = \emptyset$ and also an $r_2 >0$ such that $B(x,r_2) \cap Y= \emptyset$, but again (it's just the first argument in disguise) we take $r = \min(r_1, r_2)$ and note that $B(x,r) \cap (X \cup Y)= \emptyset$ and so $x \notin \overline{X \cup Y}$, showing that inclusion by contrapositive. For the other inclusion we just need to note that $X \subseteq X \cup Y$ implies $\overline{X} \subseteq \overline{X \cup Y}$ and similarly $\overline{Y} \subseteq \overline{X \cup Y}$ etc. From two sets we can then do finitely many by induction as you say. But I'd use the first approach plus duality.