The following is an example that I've come up with:
Suppose that $p\in A$ and $q\in B$ so that $p,q \in A\cup B$, where $A$ and $B$ are two mutually disjoint, convex, unit circles centered at $x=0,2$ in $\mathbb{R^2}$, respectively. Also let $p:=(\frac{1}{2},0)$ and $q:= (\frac{3}{2},0)$. The set of points satisfying $\lambda p + (1-\lambda)q$ for $0 < \lambda < 1$ forms a line between $p$ and $q$. But for $\lambda = \frac{1}{2}$, we have that $z = \frac{1}{2}p + (1-\frac{1}{2})q = \frac{1}{2}(p+q)=(1,0)$, which is not in $A\cup B$.
I was wondering if there's a simpler example that shows that the union of two convex sets does not have to be convex?
$(0,1) \cup (2,3)$ is a simpler example. $\frac {0.5+2.5} 2$ does not belong to this union.