Given two disjoint closed sets, prove that there exist disjoint open sets containing the closed sets Given two disjoint closed sets, $F_1$ and $F_2$ in $\mathbb{R}$, prove that there exist disjoint open sets $G_1$ and $G_2$ such that $F_1 \subset G_1$ and $F_2 \subset G_2$  

The answer is not trivial in that the distance between $F_1$ and $F_2$ could be going towards zero.
I'm outlining a proof from Baby Rudin. Let $X$ be any metric space an $\delta_A:X\to \Bbb R$ be defined by 
$$
\delta_A(x)=\inf\{d(x,a)\,|\,a\in A\}
$$
where $A$ is a subset of $X$. You can verifiy that


  
  $\delta_A$ is a continuous function on $X$ (actually, it's even uniformly continuous).
  If $A$ is a closed set, then $\delta_A(x)=0$ iff $\ x\in A$.
  


Now for any disjoint closed set $A,B \subset X$, we define $\mu:X\to[0,1]$ by
$$
\mu(x)=\frac{\delta_A(x)}{\delta_A(x)+\delta_B(x)}
$$
Verify that


  
  $\mu$ is well-defined on $X$, i.e. $\delta_A(x)+\delta_B(x)\ne 0$ all $x\in X$(Hint: Use that fact that $A$ and $B$ are closed and disjoint.)
  $0\le\mu(x)\le 1$ all $x\in X$.
  $\mu(a)=0$ for all $a\in A$ and that $\mu(b)=1$ for all $b\in B$.
  


Since $\delta_A,\delta_B$ are continuous, $\mu$ is also continuous so that $\mu^{-1}[0,\frac 12)$ and $\mu^{-1}(\frac 12,1]$ are both open sets in $X$ and that they are disjoint. It's not hard to see that $A\subset\mu^{-1}[0,\frac 12)$, $B\subset\mu^{-1}(\frac 12,1]$.

Now let $A=F_1, B=F_2, G_1=\mu^{-1}[0,\frac 12)$ and $G_2=\mu^{-1}(\frac 12,1]$ and we're done.
Such a space is called T4 space.
Not all space is T4 space, so there should be some assumtion about the space.