I'm not sure if this should be an answer or a comment, since it is an idea, or "sketch of proof ".
Consider the equivalent definition: $X$ is a simply connected space if it is path-connected and for all $x\in X$, any loop with basepoint $x$ (i.e. continuous $\alpha : I \rightarrow X$ such that $\alpha (0)=\alpha (1)=x$) is homotopy equivalent to the constant path $x$.
Suppose $U$ is not simply connected and take a simple closed curve (this is a loop), not homotopy equivalent to a constant path. This curve contained in $U$ divides the plane in two regions $A$ and $B$, one of them bounded, say $A$. The interior of $A$ has points of $U^c$, otherwise you could shrink the loop to a point staying within $U$. Observe that $B\cap U^c \neq \varnothing$ as well, since $U$ is bounded.
Now to show that $U^c$ is not connected, consider the pair of disjoint nonempty open sets of $U^c$ given by taking intersection of $U^c$ and the interiors of $A$ and $B$.