[Math] Proof that any finite subset of a metric space is closed

Question

I have a metric space $(X, d)$ and I am trying to prove that any finite subset $F = \{x_1,\ldots,x_n\} $ of $X$ is closed.
What I have by now is a proof that a subset $F$ of a metric space $X$ is closed if and only if it contains all of its accumulation points. What I think is that if I prove that my set $F$ contains all of it's accumulation points, then $F$ would be closed, right? But I have problems in prooving that $F$ contains all of its accumulation points. If anyone could tell me if I am correct and help with the last proof, that would be great.

Answer 1

Assuming that the $x_i$'s are different one from each other, take the minimum of $d(x_i,x_j)$ for $i\not=j$, noted $\varepsilon$. You obviously have $\varepsilon > 0$. Now take $\eta = \frac{\varepsilon}{2}$. Let $x$ be a limit of sequence $(u_k)_k$ of points of $F$. The sequence is convergent so it's Cauchy : there exists an $N$ such that when $k,l\geq N$, $d(u_k,u_l)\leq \eta$. Then if $k\geq N$ we have $d(u_k,u_N)\leq \eta$. But $u_N$ and the $u_k$'s are point of $F$, so by choice of $\eta$, $u_k = u_N$ for $k\geq N$, and therefore $x = u_N \in F$. $F$ is stable by convergence, so it's closed.