A finite set is closed Question:

Prove that a finite subset in a metric space is closed.

My proof-sketch: Let $A$ be finite set. Then $A=\{x_1, x_2,\dots, x_n\}.$ We know that $A$ has no limits points. What's next?
Definition: Set $E$ is called closed set if $E$ contains all his limits points.

Context: Principles of Mathematical Analysis, Rudin

If we show that the compliment of $A$ is open, then $A$ is closed.
Let $B=A^c$, and suppose that $y \in B$. We need to show that there is an $r >0$ for which $B_r(y) \cap A = \emptyset$, that is $B_r(y) \subset B$.
If we let $r = \min \{ d(y,x_i) : i = 1,...,n\}$ then it must be true that $B_r(y) \cap A = \emptyset$, since for every $z \in B_r(y)$ we have $d(z,y) < d(y,x_i)$ for each $i=1,...,n$.
