prove that any finite set in a metric space is compact

There are two ways to look at this.

One is this: suppose that we have a sequence from this finite set. Then, at least one of the values must appear infinitely many times, because if each value appears only finitely many times, then the sequence itself would be a finite number of values appearing finitely many times, which would make it a finite sequence, a contradiction. Hence, one of the values appears infinitely many times. Then, take the subsequence formed by this value appearing infinitely many times. Of course, this converges. Hence, the finite set is sequentially compact, hence compact.

The other way is even simpler: suppose we have an open cover. Then, each point is contained in some open set from the cover depending upon that point. This means there is a finite subcover (infact, the size of the subcover is at most the size of the set). Hence, the set is compact.

Both methods would do.

In metric spaces in general, being closed and bounded is not equivalent to being compact. A set $S$ is compact iff whenver $F$ is an open family (a family of open sets) such that $S\subset \cup F,$ there is a finite $G\subset F$ such that $S\subset \cup G.$

A finite set is compact.Proof by induction.

(1). If $S=\emptyset$ and $F$ is any open family then $S\subset \cup F$ and we may let $G=\emptyset.$

(2). Suppose that $n\geq 0$ and that every open cover of any $n$-member set has a finite subcover. Then if $S=\{x\}\cup T$ has $n+1$ members with $x\not \in T,$ and $F$ is an open cover of $S,$ then $F$ is also an open cover of $T,$ and $T$ has $n$ members. So there exists a finite $G^*\subset F$ such that $\cup G^*\supset T.$

Now there exists $f\in F$ with $x\in f$ because $F$ covers S. And for any such $f,$ the set $G=\{f\}\cup G^*$ is a finite subset of $F ,$ and $G$ covers $S.$