If the edge $xy$ is a bridge of $G$, then $x$ and $y$ are in separate components of $G$-$xy$.
I can't think of a counterexample so I am operating under the impression that it can be proved. How does one approach this proof? Can it be done by assuming $G$ has a cut vertex and then showing a contradiction?
A bridge is a cut-edge. So $G - xy$ separates $G$ into two separate components. So $x$ must be on one component and $y$ on the other. We want to use the fact that an edge is a bridge if and only if it is not contained in any cycle of $G$.
You could prove this by contradiction. Suppose $xy$ is a cut edge but $x, y$ are on the same component of $G - xy$. Then there exists an $x-y$ path on a component of $G - xy$. So $G$ has at least two $xy$ paths. Thus, $xy$ is contained in a cycle of $G$, contradicting the fact that it is a bridge.