Let X a set not empty and (X,d) a metric space, prove the intersection of a family of closed sets are closed.

Good morning, I'm working with this exercise. Can someone help me with that?

Let $X$ a set not empty and $(X,d)$ a metric space, prove the intersection of a family of closed sets are closed.

My proof:

Let $A_i$ with $1\leq i \leq n$ a closed sets.
We only need to prove $X-\{\bigcap_{1\leq i\leq n} A_i\}$ is open set. Let $x\in X-\{\bigcap_{1\leq i\leq n} A_i\}$, then $x\in X$ and $x \notin \bigcup_{1\leq i \leq n}A_i$ in other words, $x \in X$ and $x \notin A_1$ or $x \notin A_2$... or $x \notin A_n$
As $A_i$ are closed sets, then, $X-\{A_i\}$ are open sets. $(*)$
Suppose $x\notin A_i$, then for $(*)$ exists $r>0$ such that $B(x,r)\subset X-\{\bigcap_{1\leq i \leq n}A_i\}$
In consequence, $X-\{\bigcap_{1\leq i\leq n} A_i\}$ is open set.

Is correct the proof? Is convincing?

You can also define the closure of a set $A$ as $\{x \in X \mid \forall r>0: B(x,r) \cap A \neq \emptyset \}$, the set of adherence points of $A$. A set is closed iff it equals its closure.
So let $C_i$ ,$i \in I$ be a collection of closed sets and define $C = \bigcap_{i \in I} C_i$. Let $x$ be an adherence point of $C$, but suppose (for a contradiction) that $x \notin C$. This means that there is some $i_0 \in I$ such that $x \notin C_{i_0}$. So $x$ is not an adherence point of $C_{i_0}$ (as $C_{i_0}$ is closed), so there is some $r_0 > 0$ such that $B(x,r_0) \cap C_{i_0}= \emptyset$, and a fortiori we have that this $r_0$ shows that $x$ is not an adherence point of $C$, as $C \subseteq C_{i_0}$, contradiction. So $C$ is closed, as it equals its set of adherence points.