Compact sets in Hausdorff spaces. 
$(X, \mathcal{T})$ is a topological spaces. $X$ is Hausdorff. $A,B$ are compact, disjoint sets in $X$. 

Suppose $X$ is Hausdorff and $A \subset X$, $B \subset X$ are compact and disjoint. Consider an arbitrary point $a_i \in A$. Since $A \cap B = \emptyset$, $a_i \in X - B$. Let $b_i \in B$. Clearly, $a_i \neq b_i$. Since, $X$ is Hausdorff, there exists an open neighbourhood $U(a_i)$ of $a_i$ and an open neighbourhood $V(b_i)$  of $b_i$ such that $U(a_i) \cap V(b_i) = \emptyset$. \ The set $\mathcal{V} = \{V(b_i) \mid b_i \in B\}$ is an open cover of $B$. By compactness of $B$, there exists a finite subcover of $B$ contained in $\mathcal{V}$, say $\{V(b_1),V(b_2), \dots ,V(b_n)\}$. It follows that $B \subset \bigcup_{i=1}^n V(b_i)= V$. Clearly $V \in \mathcal{T}$. Similarly, there exists $U \in \mathcal{T}$ such that $A \subset \bigcup_{i=1}^n U(a_i) = U$. 

Now, how do I show $U \cap V = \emptyset$ ?

 A: I think what the questioner is after is a Corollary of the following result which is 3.5.6 on p.84 of the book Topology and Groupoids. 
3.5.6 Let $B,C$ be compact subsets of $X,Y$ respectively and let$\mathcal W$ be a cover of $B \times C$ by sets open in $X \times Y$. Then $B,C$ have open neighbourhoods $U,V$ respectively such that $U \times V$ is covered by a finite number of sets of $ \mathcal W$.
The proof starts with the case $B$ is a singleton,  and then proceeds to the general case. 
Corollaries of this are:


*

*The product of compact spaces is compact. 

*Let $B,C$ be compact subsets of $X,Y$ respectively and let $W$ be an open subset of $X \times Y$ containing $B \times C$. Then $B,C$ have open neighbourhoods $U,V$ respectively such that $U \times V \subseteq W$. 

*If $B,C$ are disjoint compact subspaces of the Hausdorff space $X$, then $B,C$ have disjoint open neighbourhoods.

*A compact subset of a Hausdorff space is closed. 
Edit March 1: Actually there are some lessons for a beginner to derive from the answers and comments. 
It is almost always useful  to look for the simplest possible case to get started. Here one first tries for both $A,B$ singletons; but that is no good because the result is just the usual Hausdorff condition. So one next tries $A$ compact and $B$ a singleton, and this, as the solutions show, gets one started. 
Another method is to  consider other possible outlooks on the problem, even if one already has a solution. Thus a condition equivalent to the usual Hausdorff condition on $X$ is that the diagonal $$\Delta_X= \{(x,x)| x \in X\} \;$$ is closed in $X \times X$. Further $A,B$ disjoint is equivalent to $(A \times B) \cap \Delta_X= \emptyset$. Putting these together, one is led to something like a special case of 2. above. 
A: I have a proof. I describe intuitively here.
Pick $a\in A$, for each $b\in B$ we can find $a\in U_b$, $b\in V_b$ open such that $U_b,V_b$ disjoint. The set of all $V_b$ is an open cover of $B$, we find a finite subcover of $V_{b_k}$, and we take intersections of $U_{b_k}$, it a open neighborhood of $a$ such that it does not intersect the union of $U_{b_k}$. Do so for every point in $A$, we will get again an open cover of $A$, and obtain a subcover from it. Take the intersection of the subcover of $A$ to be the final open neighborhood, and the union of corresponding finite subcover of $B$ to be the final open neighborhood, they do not intersect and caontain $A$ and $B$ correspondingly.  
