Does a Vertex Cover exist?
If you delete the vertex $v$, then there are no edges left at all, and so there is a vertex cover of size $0 \leq k-1$.

Why is the statement true in the first place? Suppose that $G$ has a vertex cover $C$ of size $k$. Since $(u,v)$ is an edge, $C$ must contain at least one of $u,v$, say $v$. I claim that $C \setminus v$ is a vertex cover of $G \setminus v$, from which the statement immediately follows. Indeed, let $(a,b)$ be any edge of $G \setminus v$. Then $(a,b) \in G$, and so $C$ contains one of $a,b$, say $a$. By construction, $a \neq v$, and so $a \in C \setminus v$.