Every open neighborhood of a point in the closure contains a point in the set Definitions:

Neighborhood: The neighborhood of a point $x$ is a subset of the topological space $X$ that contains an open set such that $x$ is in that open set.
Closure: The closure of a set $S$ is the intersection of all the closed subsets of the topological space $X$ which contain $S$ as a subset.
Closed: A set is closed if the complement of that set is open.

I want to prove that every open neighborhood of a point $x$ in the closure of a set $S$ contains a point in $S$.
If $x \in S$, then we're done. If $x \not\in S$, then $x \in \operatorname{cl}S \setminus S$. Since we don't know if $S$ is open, closed, both, or neither, I guess we have to do cases:
1) $S$ is closed: then $S = \operatorname{cl}S$, so we're done.
2) $S$ is open: then $S = \operatorname{Int}S$ and $\operatorname{cl}S \setminus S$ is closed. If $x \in \operatorname{cl}S \setminus S$, and there exists an open neighborhood of $x$.... I don't know how to show that a point in this neighborhood can also be in $S$.
3) $S$ is both open and closed: No idea how to start this one
4) $S$ is neither open nor closed: No idea how to start this one
 A: If $S$ is open and closed, then you can use either argument $1$ or argument $2$, since argument $1$ does not say "if $S$ is closed, but not open".
That said, I think your separation into cases is not the optimal way of going about this. You are better of simply taking $x\in\mathrm{cl}(S)$ and a neighborhood $O$ of $x$ and show that $O\cap S\neq \emptyset$.
For that, you can assume that $O\cap S = \emptyset$, then notice how the complement of $O$ is a closed set which contains $S$, meaning that it also contains the closure of $S$. Can you come up with a contradiction now?
A: We proceed by contraposition and prove: if a point $x$ has a neighbourhood  (not necessarily open) that doesn't meet $S$, it is not in the closure of $S$.
Indeed, we may suppose this neighbourhood $V$ is open. Then, $F_X\smallsetminus V$  is a closed subset of $X$, and it contains $S$. Hence $x$ doesn't lie in all closed subsets of $X$, i.e. it doesn't lie in the closure of $S$.
Note. This proof is valid in any topological space, not only in metric spaces.
A: We have to show that $B(x,r)\cap S\neq \emptyset \forall r>0$
Suppose $B(x,r)\cap S= \emptyset $ for some $ r_1>0$
Then $S\subset X\setminus B(x,r_1)$ ; $X\setminus B(x,r_1)$  is a closed set not containing $x$.
But we have assumed that $x\in S_\alpha$ forall $\alpha$ where $S_\alpha$ is a closed set containing $x$ 
Thus arriving at a contradiction because we have got  a closed set not containing $x$
