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

Note that $A = \bigcup_{i=1}^n \{x_i\}$. Also each $\{x_i\}$ is closed since $\Bbb R \setminus \{x_i\} = (-\infty,x_i)\,\cup (x_i, \infty)$ and both $(-\infty,x_i)$, $(x_i, \infty)$ are open in usual topology on $\Bbb R$ (You can also check this without invoking topology by considering the fact that any point in any of these subsets has a open neighbourhood lying in that subset and hence, every point of these subsets is an interior point. So, both of these are open).
Also, we know that the finite union of closed sets is closed and hence every finite set is closed in a metric space.  
