Hello
Suppose $\displaystyle x$ is an accumulation point of a set $\displaystyle A$. Then there exists distinct $\displaystyle x_n \in A$ such that $\displaystyle x_n \to x$. Let $\displaystyle r>0$ then by definition of convergence it means $\displaystyle d(x,x_n) < r$ for $\displaystyle n\geq N$. Thus, $\displaystyle x_N,x_{N+1},...$ are distinct elements which lie in $\displaystyle B(x,r)$ and hence there are infinitely many.
A set is closed if and only if it contains all its accumulation points. A finite set clearly has no accumulation points, hence the set is empty. And it contains the empty set.Using this, prove that any finite subset of X is closed.