If two sets are separated, then any two subsets of those sets are also separated? I want to prove that if two sets X and Y are separated, then subsets of those sets are also separated.
The definition is that if X intersect Y closure is empty and X closure intersect Y is empty, the sets are separated.
So can I say that if no point in X lies in the closure of Y, then no point in any subset of X lies in the closure of Y, and that if no point in Y lies in the closure of X then no point in any subset of Y lies in the closure of X? 
Would that be enough to prove the claim?
Thank you in advance!
 A: Your reasoning is correct, but it's a bit disorderly. Mainly, I think what you ought to note is that, using $\text{cl}$ as the closure operator that if $A\subseteq X$ and $B\subseteq Y$ then $\text{cl}(A)\subseteq\text{cl}(X)$ and $\text{cl}(B)\subseteq \text{cl}(Y)$. Clearly, if two sets (e.g. the closure of $X$ and $Y$) are disjoint, then so are two subsets of those sets. 
This just makes it more explicit that we are talking about a pair of subsets $A$ and $B$ of $X$ and $Y$ respectively (rather than leaving them unnamed, which makes it a lot wordier and confusing). Also, it's not really necessary to go through every little case of "$\text{cl}(A)$ is disjoint from $Y$" and "$\text{cl}(B)$ is disjoint from $X$" as you do, since the proofs are entirely analogous - just operating on different symbols - and it's generally better not to repeat yourself.
A: Sounds like you have the right idea. More formally you might say:
Let $X,Y$ be separated sets with subsets $A,B$ respectively. Since $$A\cap B \subset X \cap Y = \emptyset$$ then $A\cap B = \emptyset$.
