This is not true in general. If a space is first countable, i.e. has a countable basis at each point, then this result is true. Metric spaces are first countable.
To prove this, suppose $X$ is a first countable space and $A \subset X$. (In your case, $A = \{x_n : n \in \mathbb{N}$ where $(x_n)$ is a sequence.) Let $(U_n)_{n \in \mathbb{N}}$ denote a countable basis at the point $x^*$, where $x^*$ is an accumulation point of $A$. By definition, of accumulation point, there exists a $x_0 \in A$ and $x_0 \in U_0$. $U_0 \cap U_1$ is an open set. Again since $x^*$ is an accumulation point, there exists a $x_1 \in A$ and $x_1 \in U_0 \cap U_1$. Continue this to create a sequence $(x_n)_{n \in \mathbb{N}}$. Then since $U_n$ is a countable basis, you can prove that $\lim_{n \rightarrow \infty} x_n = x^*$. This is because given any open set $U$, there exists a $n$ such that $U_n \subset U$ such that $x^* \in U_n$ (definition countable basis). By construction, for all $m > n$, $x_m \in U_n \subset U$.
In metric spaces, the proof is bit nicer. You can take your countable basis to be $U_n = B_{\frac{1}{n}}(x^*)$. So using the definition of accumulation point, just choose $x_n$ such that $x_n \in B_\frac{1}{n}(x^*)$.
An accumulation point of a sequence is a more general concept that a limit. For example, the sequence $+1,-1,+1,-1,...$ has no limit, but two accumulation points, $\pm 1$.
(If you prefer to have distinct points, take the sequence $+1+\frac{1}{1},-1+\frac{1}{2},+1+\frac{1}{3},-1+\frac{1}{4},...$)
Think of accumulation points as limits of subsequences. A point is an accumulation point of a sequence iff you can find a subsequence converging to that point.
It should be clear that if $a_n \to A$, then all subsequences must also converge to $A$.
Suppose $J$ is an infinite set, and $U$ an open set containing $A$. Since $a_n \to A$, we have some $N$ such that $a_n \in U$ for all $n \ge N$. Then we see that the set $J'=J \cap \{N,N+1,...\}$ is also infinite (otherwise a quick contradiction), and for all $n \in J'$, $a_n \in U$. Hence $A$ is an accumulation point of the subsequence $a_n$, $n \in J$.
Best Answer
Assume that $x$ is an accumulation point of $x_n$. Then every neighborhood of $x$ contains a point of $x_n$. Let $m>0$. Then $B(x,m)$, the open ball centered at $x$ with radius $m$, is a neighborhood of $x$ and hence contains a point of $x_n$. Repeat this process for $m/2, m/4, \ldots$ to obtain a subsequence of points contained within $B(x,m/2)$, $B(x,m/4)$, $\ldots$. Then $B(x,m)$ contains all of these points, of which there are infinitely many, as was to be proved.
Assume now that for each $m>0$, there are infinitely many $n$ such that $|x_n-x|<m$. Let $U$ be a neighborhood of $x$. Then there exists $\varepsilon>0$ such that $(x-\varepsilon,x+\varepsilon)\subset U$. Choose $m<\varepsilon$, then there is an $n$ such that $|x_n-x|<m<\varepsilon$ and hence $x_n\in U$. It follows that $x$ is an accumulation point of $x$.