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
Thinking $A$ as a subset of a metric space $M$. An easy approach will be to use that the single points $\{x_j\}\subset M$ are closed (you know why?), then of course
$$
A=\bigcup_{j=1}^n \{ x_j \}
$$
Since $A$ is a finite union of closed sets, it is itself closed.
If some open set $G_i$ containing $x_1$ does not contain $x_i$, for $i=1,\ldots,n$, then $\bigcap_{i=1}^n G_i$ is an open set containing $x_1$ and not containing any of $x_2,\ldots,x_n$.  Therefore $x_1$ is not a limit point of the set.  The same argument applies to show that $x_2,\ldots,x_n$ are not limit points.

If the space is a Hausdorff space, then certainly some open set $G_i$ containing $x_1$ does not contain $x_i$ if $i\ne1$.

If it's a metric space, you can take $G_i$ to be the open ball of radius $d(x_1,x_i)$ centered at $x_1$.