Problem: $G$ is a bipartite graph having partite sets $U$ and $W$ where $|U|$ = $|W|$ = $k \geq 2$. I need to prove that if every two vertices of $U$ have distinct degree in $G$, then $G$ contains a perfect matching.
Attempt at Proof: (I believe it's complete except the annoying case where we have an isolated vertex). The following proof assumes that no vertex in $U$ is an isolated vertex and proving one cannot exist is what I'm struggling with
If all pairs of vertices of $U$ have different degree, then every vertex in $U$ has different degree. Any subset $S \subseteq U$ has some vertex $x$ with largest degree, and so the size of $N(S)$ is at least $deg(x)$, as all vertices in $S-x$ have neighborhoods that are proper subsets of $N(x)$. In other words, the $x$ dictates how many vertices $N(S)$ must have at the least, and this number will always be at least $|S|$ by our assumption that every vertex has distinct degree and no vertex has degree $0$.
Therefore, every subset $S \subseteq U$ meets Hall's bipartite condition so there exists a matching that saturates every vertex of $U$, but since $|U| = |W|$, such a matching saturates every vertex in $G$ and is thus perfect.
This should work, but how do I get rid of this problematic case of isolated vertices? They violate Hall's condition and make things harder, but there's no straightforward way to rule them out with the assumptions we are given.
Notation: $N(S)$ where $S$ is a set is the union of the neighborhoods of each vertex $s \in S$.
Edit: I didn't read the question carefully, it said that $G$ was a connected bipartite graph. This solves the problem as a connected bipartite graph with more than 1 vertex cannot contain isolated vertices.