Removing finite points from closed set. I have the following homework problem:

"Let $S$ be a closed set and let
$$
S_1 = \{x_1, x_2,..., x_n\} \subset S
$$
Is $S \smallsetminus S_1$ also a closed set? If true, prove the statement. If false, give a counterexample".

I think I have a counterexample to the statement, but I'm not completely sure of its validity:

Let $S =\{p\} $ a set composed of a single point. Since $S$ contains only one point, $S$ is closed. Let $S_1 =S$. Since any set is a subset of itself, then it holds true that $S_1 \subset S$. We then have:
$$
S \smallsetminus S_1 = S \smallsetminus S = \emptyset
$$
Since $ \emptyset$ is an open and closed set, in particular, it is open. Hence the statement is false.

My question is if it's valid to just assume that the empty set is in particular open and "ignoring" that it's also closed. Does this counterexample hold?
No, your example doesn't work because the empty set is closed. Being open does not prevent a set from being closed, as they are not mutually exclusive conditions. However you could just remove $0$ from the interval $[0,1]$ to obtain a counterexample.

This works if you want to show it's not true in general. More generally, if there exists a one point set in the space that is not open, then the whole space minus that point is a counterexample. Thus the discrete topology is actually the only situation where the statement holds.
Recall that in order for a set to be closed, it must contain all of it's accumulation points. 

Well, a point $x$ of a set $A$ is said to be an accumulation point if for any neighborhood of $x$, the neighborhood contains at least one other point in the set $A$.

We know that any finite set is clearly not infinite. So, you can generate an extremely small neighborhood about any point in a finite set and get infinitely many points not in the original set.

Your question doesn't specify if this set is infinite or finite. Assuming finite $S$, then yes, any $S\setminus S_1$ would contain no accumulation points and be closed.

If $S$ is infinite, closed, and you remove any part it, you're removing accumulation points and thus ensuring the set is no longer closed. However, this is only true unless the infinite set has no accumulation points to begin with. 

Another problem arises with the topology being used here. It's not specified, but in a general case this is the structure.