A tree edge $uv$ with $u$ as $v$’s parent is a cut edge if and only if there are no edges in $v$’s subtree that goes to $u$ or higher
We must prove two directions: (i) if $uv$ is a cut edge, then there is no edge from $v$'s subtree to $u$ or higher and (ii) if there is no edge from $v$'s subtree to $u$ or higher, then $uv$ is a cut edge.

For (i), if there was an edge from $v$'s subtree to $u$ or higher, then the removal of $uv$ would not increase the number of connected components contradicting the fact that $uv$ is a cut edge. To prove (ii), since every path from any vertex in a subtree of $v$ to a vertex that is $u$ (or higher than $u$) must use the edge $uv$, we see that $uv$ is indeed a cut edge.