Prove that $G - F$ is open and $F - G$ is closed given $F$ closed and $G$ open in $M$. 


proof:
Since  $G$ is open, $\exists B_{r}(x) \subset G$ for any $x$ and for any $r > 0$, $B_{r}(x) \cap F = \emptyset.$ So $B_{r}(x) = B_{r}(x) - F \subset G - F.$ (I am not sure how to justify $B_{r}(x) - F \subset G - F$…the venn diagram provided the intuition, is it necessary to justify this ?)
Since $F$ is closed, $\forall x \in F$, $B_{\epsilon }(x) \cap F = \emptyset$ and $B_{\epsilon}(x) \not\subset G$ for any $\epsilon >0.$ So $B_{\epsilon}(x) \cap (F - G) =B_{\epsilon}(x) \cap F - B_{\epsilon}(x) \cap G = B_{\epsilon}(x) \cap F \neq \emptyset.$
I am sure the last part proving $F-G$ is right.

You have to be more careful with your quantifiers.

Statement: $G-F$ is open.
We want to prove that, for every $x\in G-F$, there exists $r>0$ such that $B_r(x)\subset G-F$.
So, let $x\in G-F$. In particular $x\in G$ and $x\notin F$.
Since $G$ is open, there is $s>0$ such that $B_s(x)\subset G$.
Since $F$ is closed, there is $t>0$ such that $B_t(x)\cap F=\emptyset$.
Now, take $r=\min\{s,t\}$. Then $B_r(x)\subset B_s(x)\subset G$ and $B_r(x)\subset B_t(x)$ so $B_r(x)\cap F=\emptyset$. Thus $B_{r}(x)\subset G-F$.

Statement: $F-G$ is closed.
We want to prove that, for every $x\notin F-G$, there exists $r>0$ such that $B_r(x)\cap (F-G)=\emptyset$.
So, let $x\notin F-G$.
If $x\notin F$, take $r>0$ such that $B_r(x)\cap F=\emptyset$ (it exists since $F$ is closed); then, easily, $B_r(x)\cap (F-G)=\emptyset$.
If $x\in F$, then we conclude that $x\in G$, because otherwise $x\in F-G$.
Thus there is $r>0$ such that $B_r(x)\subset G$ (because $G$ is open). Now it's easy to see that $B_r(x)\cap (F-G)=\emptyset$.
