Definition. A Graph is almost $r$-regular if each vertex has degree $r-1$ or $r$.
Theorem. Let $G$ be almost $r$-regular for $r\geq 2$, then $G$ contains an almost $(r-1)$-regular graph $H$ with $V(H)=V(G)$.
Idea of proof. Remove any edges that can be removed while keeping $G$ almost $r$-regular. Partition $V(G)$ in $A$ and $B$ s.t. vertices of degree $r$ are in $A$ while vertices of degree $r-1$ are in $B$. Note that no edge has both endpoints in $A$. If we find a matching from $A$ to $B$, we are done by removing all matching edges. So for each $S\subseteq A$ we want $|S|\leq|\Gamma(S)$| (Hall's condition).
Problem. By double-counting the edges between $S$ and $\Gamma(S)$ I only get $r\cdot|S|\leq(r-1)\cdot|\Gamma(S)|$. It seems reasonable that Hall is satisfied but I'm struggling to make it explicit.
You are really only missing a very small last insight.
Disregarding the edges "inside" $\Gamma(S)$, every vertex of $S$ has degree $r$, so $r|S|$ edges from $S$ arrive at $\Gamma(S)$. Since each vertex of $\Gamma(S)$ can only accommodate at most $r-1$ edges, $\Gamma(S)$ must have more than $|S|$ vertices.
So Hall is satisfied and you are done.