If $F$ is closed and $x \not\in F$, show that there are disjoint open sets $U,V$ such that $x \in U$ and $F \subset V$ If $F$ is closed and $x \not\in F$,

(a) show that there are disjoint open sets $U,V$ such that $x \in U$ and $F \subset V$.

(b) Can $U$ and $V$ be chosen such that their closures $\bar{U} , \bar{V}$are disjoint as well ?

Attempt:
Assume $d(x,F) = t$

Consider the function $d(y,F)$ which is continuous, so the set $A = \{y | d(x,F) &gt; t/2\}$ is open and contains $x$, similarly the set $B = \{y | d(x,F) &lt; t/2\}$ is also open and contains F, both A and B are disjoint hence the required sets.

Can I consider the closure of the A and B for the second part (b) of the question?
One missing piece in this proof: You need to know that $t&gt;0$ in order to prove that $A,B$ are open. So, for that purpose, you need to prove that $t&gt;0$. 

Regarding part (b), the closure of $A$ and $B$ might, by bad luck, contain the same point $z$, such that $d(z,F)=t/2$. However, you should be able to rewrite your proof to avoid that bad situation and obtain disjoint closures.
Let r =d(x,F)/100
Now we define U=B(x,r) and V=UB(a,r)
a€F
Clearly, x€U and F is contained in V.
Then we claim that U and V are disjoint.
If not, then let there exists z€UnV.
=&gt; d(x,z)&lt;r and d(z,y) &lt; r
⇒ d(x,y) &lt; 2r= d(x,F)/50,which is a contradiction .
Hence our claim &quot;U and V are disjoint &quot; is true. Q.E.D