Note that $v$ is a leaf in the spanning tree given by BFS($v$) if and only if $v$ is a leaf in $G$ (by definition of BFS).
That said, you should try with a graph $G$ that is a circular graph (or cycle graph), and any vertex $v=s$.
If you want an answer when $v \neq s$, then we can indeed show that if $v$ is a leaf in any DFS($s$), then $v$ is a leaf in $G$ (and so $v$ is a leaf in any BFS($s$)).
Indeed, suppose $v$ is not a leaf in $G$. Then it means that $v$ has two neighbors $x$ and $y$. If we suppose that $v$ is a leaf in any DFS($s$), then consider such a DFS where $x$ is explored before $y$. We have now two possibilities:
- $v$ is explored before $x$. Since $v\neq s$, that means that $v$ is not a leaf in the resulting spanning tree (because in this tree, $v$ has two neighbors);
- $v$ is explored after $x$. Then a graph traversal similar to this one until $x$, which then explores $v$ then $y$ then continues the DFS from $y$ would still be a DFS. In the resulting spanning tree, $v$ has two neighbors, so $v$ is not a leaf.
We proved that $v$ not a leaf in $G$ implies there exists a DFS($s$) where $v$ is not a leaf, or equivalently $v$ is a leaf in every DFS($s$) implies $v$ is a leaf in $G$ implies $v$ is a leaf in every DFS($s$).