I read the proof of the Tutte-Berge formula from the book "Fundamentals of Graph Theory" by Allan Bickle (page 204), but one detail I did not understand.
For a graph to have a 1-factor, Tutte's condition must hold, since each odd component must have some vertex matched to a vertex in $S$. More generally, any matching must miss $n-\operatorname{def}(G)$ vertices, so the size of a maximum matching is at most $\frac{1}{2}(n-\operatorname{def}(G))$. Tutte proved the converse of the first statement, characterizing graphs with 1-factors. Berge extended this result to prove that $\alpha^{\prime}(G)=\frac{1}{2}(n-\operatorname{def}(G))$
Theorem $7.20$ (Tutte-Berge Formula-Berge [1958]). If $G$ is a multigraph, then $\alpha^{\prime}(G)=\frac{1}{2}(n-\operatorname{def}(G))$.
Proof (Schrijver [ [2003] $)$. We have noted that $\frac{1}{2}(n-\operatorname{def}(G))$ is an upper bound. We prove the reverse inequality by induction on $n$. The case $n=1$ is trivial. We can assume that $G$ is connected, as otherwise we can apply induction to the components of $G$.
First assume that there exists a vertex $v$ covered by all maximum-size matchings. Then $\alpha^{\prime}(G-v)=\alpha^{\prime}(G)-1$, and by induction there exists a subset $U^{\prime}$ of $V(G)-v$ with $$\alpha^{\prime}(G-v)=\frac{1}{2}\left((n-1)+\left|U^{\prime}\right|-o\left(G-v-U^{\prime}\right)\right). $$ Then $U=U^{\prime} \cup\{v\}$ gives equality. So we can assume that there is no such $v$. In particular, $\alpha^{\prime}(G)<\frac{n}{2}$. We show there is a matching of size $\frac{n-1}{2}$, which implies the theorem (with $U=\emptyset$ ).
Indeed suppose to the contrary that each maximum-size matching $M$ misses at least two distinct vertices $u$ and $v$. Among all such $M, u, v$, choose them such that the distance $d(u, v)$ of $u$ and $v$ in $G$ is as small as possible.
If $d(u, v)=1$, then $u$ and $v$ are adjacent, and hence we can augment $M$ by $u v$, contradicting the maximality of $|M|$. So $d(u, v) \geq 2$, and hence we can choose an intermediate vertex $t$ on a shortest $u-v$ path. By assumption, there exists a maximum-size matching $N$ missing $t$.
Consider the component $P$ of the graph induced by $M \cup N$ containing $t$. As $N$ misses $t, P$ is a path with end $t$. As $M$ and $N$ are maximum-size matchings, $P$ contains an equal number of edges in $M$ as in $N$. Since $M$ misses $u$ and $v, P$ cannot cover both $u$ and $v$. So by symmetry we can assume that $P$ misses $u$. Exchanging $M$ and $N$ on $P, M$ becomes a maximum-size matching missing both $u$ and $t$. Since $d(u, t)<d(u, v)$, this contradicts the minimality of $d(u, v)$.
What confused me was the last paragraph of the narrative.
Consider the component $P$ of the graph induced by $M \cup N$ containing $t$. As $N$ misses $t, P$ is a path with end $t$.
I'd like to ask that the component $P$ being a path can be derived from the fact that $N$ misses $t$.
In the graph induced by $M \cup N$, every vertex has degree at most $2$. You should convince yourself that this means that every component of this graph is either a cycle or a path. Also note that $M$ does not miss $t$, since if it did, that would contradict the choice of $u$ and $v$. Since $N$ misses $t$, the degree of $t$ in the graph induced by $M \cup N$ is exactly one. It follows that its component must be a path, and $t$ must be an endpoint of this path.