Question: I understand that Lebesgue outer measure on $\mathbb R$ is not countably additive. But if there are two disjoint sets, does the outer measure of their union equal the sum of their outer measure? Can someone give me a counterexample?

Answer: 
Take $\Omega = \{a, b\}$. Define
$$
 \mu^\*(A)
 =
 \left\{
 \begin{array}{ll}
 0, & A = \emptyset
 \\
 1, & A \neq \emptyset
 \end{array}
 \right.
$$
This is an outer measure.

In this case,
$$
 \mu^\*(\{a,b\}) = 1
 \neq
 2 = \mu^\*(\{a\}) + \mu^\*(\{b\}).
$$
For a non-measurable set, the outer measure will necessarily be positive. Therefore it suffices to consider a non-Lebesgue measurable subset $A$ of the circle, and let $B$ be its complement in the circle. Then additivity must fail for this pair of sets for the following example. 

Consider a set $A$ of representatives for equivalence classes under the action of rational rotations on the circle (the usual example of a non-measurable set). Recall that the circle is a countable union of the sets $A\_\alpha$ obtained by rotating $A$ by a rational angles $\alpha$.

Let $\epsilon=\mu(A)>0$ where $\mu$ is the outer measure. Let $N=1+\lfloor\frac{1}{\epsilon}\rfloor$. Now take $N$ rational translates of $A$, namely $A\_1,A\_2,\ldots,A\_N$. These are all disjoint by construction. If $\mu$ satisifed finite additivity, then the total outer measure would be greater than $1$:
$$
\mu(A\_1\cup\cdots\cup A\_N)=N\cdot\epsilon>1,
$$
yielding the desired contradiction. This shows that the outer measure is not even finitely additive.