Let $(X,d)$ be a metric space. Let $F \subseteq X$ be a finite set. Prove that $F$ is closed.
Proof. Suppose $(a_n)$ is convergent in $F$. Because every convergent sequence in a finite metric space is eventually constant, then $\exists N \in \mathbb{N} \ni \forall n \ge N$, we have $a_n=a_N \in F$. As $a_N \in F$, then $(a_n) \to a_N \in F$. Thus $lim_{a_n} \in F$, and hence $F$ is closed.
Is this correct?
And now I would like to prove this by induction on the number of points (the cardinality) in $F$.
Let $(X,d)$ be a metric space. Let $F \subseteq X$ be a finite set. Prove (by induction on the cardinality of $F$) that $F$ is closed.
What I've done so far:
Proof. Base case: If $F=\emptyset$. Then the claim is vacuously true since $\emptyset$ does not contain any sequence $(a_n)$. In other words, the assertion is false; and therefore, the implication that $lim_{a_n} \in \emptyset$ is true.
Assume that for some natural number $n \ge 0$ that $|F|=n \Rightarrow F$ is closed. Since $F$ is closed, then $(a_n) \in F$ such that $(a_n) \to x$, where $x \in F$.
Any insight would be appreciated!
Instead of an approach using sequences, it would be easier if we used the fact that finite union of closed sets is closed.