A finite set is closed

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.

I think the simplest answer to this, following Baby Rudin (Principles of mathematical analysis 3rd ed.) terminology, is:

The definition of a closed set is in 2.18.d: a set $A$ is closed if every limit point of $A$ is an element of it. However, none of the points $x_i \in A$ are limit points of $A$: for each $x_i$, pick $r = \min(d(x_{i-1}, x_i), d(x_i, x_{i+1}))$ i.e. the minimum distance between itself and its neighbours (to take care of corner cases, define $d(x_0, \_) = \infty$ , $d(\_, x_{n+1}) = \infty$). Now, any neighbourhood $N_{r'}(x_i)$ for $r' < r$ doesn't contain any other point of $A$ apart from $x_i$ so $x_i$ can't be a limit point. Therefore, $A$ is closed "vacuously" i.e. trivially, because the condition for it to be closed (all limit points of $A$ are elements of $A$) is trivially satisfied because the set of all limit points of $A$ is empty (a set of 0 points has any property). Another way to say this is that the empty set is a subset of any set.

If $M$ is a metric space then every subset $A =\{x_1, \ldots, x_n\} \subseteq M$ is closed. In fact, if $a \notin A$ then $d(a,A)$ is the least of the numbers $d(a,x_1) ,\ldots, d(a,x_n)$ thus, $d(a,A) > 0$.

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 $i=1,...,n$.