Cut edge and cut vertices
Let $G=(V,E)$. By symmetry it suffices to show that $u$ must be a cut-vertex.
Let $C$ be the set of vertices in the connected component of $G-(u,v)$ containing $u$. Clearly $v \not\in C$.
Let $x$ be a neighbor of $u$ distinct from $v$. Since $(x,u) \in E \setminus \{(u,v)\}$, we must have $x \in C$ and hence $x$ is disconnected from $v$ in $G-(u,v)$.
Since $G-u$ is a subgraph of $G-(u,v)$ containing both $u$ and $v$, $x$ must be disconnected from $v$ in $G-(u,v)$. The claim follows by noticing that $x$ was connected to $v$ in $G$.