Nice and tricky problem. I believe there's a small typo in the statement. One direction is obvious, if a graph $G$ with degree sequence $d_1$, $d_2$, $\dots$, $d_n$ contains a perfect matching, then by deleting the edges of the perfect matching, we get a new graph whose degree sequence is $d_1-1$, $d_2-1$, $\dots$, $d_n-1$. For the converse, we will restate the problem and prove the following: Quote: Let $v_1$, $v_2$, $\dots$, $v_n$ be the vertices of both a graph $G$ with degree sequence $d_1$, $d_2$, $\dots$, $d_n$ and $G'$ a graph with degree sequence $d_1-1$, $d_2-1$, $\dots$, $d_n-1$. Choose them as to share the maximal amount of edges. Then, $G'$ is contained in $G$. For easier notation, say an edge that belongs to $G$ blue, $G'$ red, or if it belongs to both purple (an edge can't be both red and purple, or blue and purple). Claim. For a quadruple of vertices $a$, $b$, $c$, $d$ exists with $ab$, $bc$, $cd$ painted with blue, red and blue respectively, then $ad$ is also blue (notice that you can swap blue and red here). Proof. Delete $bc$ and make $ab$ and $cd$ purple (need some care with two cases). $\blacksquare$

FTSOC, assume that we have red edges, then pick the vertex $v$ with maximal $M$ number of red edges incident to itself say $v_1$, $v_2$, $\dots$, $v_M$, let $w$ be a vertex incident to $v$ with edge $vw$ blue and $x$ be a vertex incident to it with edge $vx$ red. Pick a vertex $y$ from the neighborhood of $x$ with $xy'$ blue, by the claim, $wy$ is also blue, implying the existence of a vertex $z$ with $zw$ red (some of these vertices may coincide, but the procedure is the same), again by the lemma, this implies that $v_iz$ is red for $1 \le i \le M$, contradicting maximality.

Remark. Click to reveal hidden text

Originally from the book <Combinatorial problems and exercises>, §7-53.

As already mentioned by kingu, this is a particular case of Kundu's k-factor Theorem, whichs proof can be found in the attached file.

Attachments:
KunduKFactor.pdf (280kb)