Proving that $X$ is a closed subset if and only if whenever $B_\epsilon (x) \cap X \neq \emptyset $ for every $\epsilon >0$, then $x \in X$. Suppose $X$ is a subset of a metric space $M$. I would like to prove that $X$ is a closed subset of $M$ if and only if whenever $x$ is a point in $M$ such that $B_\epsilon (x) \cap X \neq \emptyset $ for every $\epsilon >0$, then $x \in X$.
I did one direction (the easier one), but am having trouble coming up with a rigorous proof for the other direction. Any help would be graciously appreciated!

Definitions I am using:
If every limit point of $X$ belongs to $X$, I say that $X$ is closed. Also, $X$ is open if and only if $X^\mathsf{c}$ is closed.
Let $M$ be a metric space and $X$ be a subset of $M$. We say that a point $x$ in $M$ is a limit point of $X$ if there is a sequence $x_n$ such that $x_n \in X$ for every positive integer $n$ and limit as n goes to infinity of $x_n = x$. 

HINT: Construct a sequence tending to $x$ by considering different values of $\epsilon$, like $\frac{1}{n}$ for all $n$ and finding a point, say $x_n$, in $B_\epsilon (x) \cap X$. And the apply definitions. This will work to prove both sides of the double implication. 
