First, you’ve reversed your conclusions: if the arguments were correct, the first would show that $\beta'(H)\le\alpha(H)$ and the second that $\beta'(H)\ge\alpha(H)$. The second argument is correct, but the first isn’t. To see where it goes wrong, apply it to $K_3$, a triangle. $X$ contains a single vertex, and both vertices in $Y$ are connected to that same vertex: your edge cover has cardinality $2$, not $1$ (and indeed $\alpha(K_3)=1$ while $\beta'(K_3)=2$).
Assume that $\alpha(H)=\beta'(H)$ for each subgraph $H$ of $G$ having no isolated vertices; you want to show that $G$ must be bipartite. HINT: If not, $G$ contains an odd cycle.
Edit, 11 Jue 2020
My original suggested proof for the other direction was nonsense, as pointed out by Mahsa in another answer while I was away and now by PAB in the comments; I tried to make it easier than it actually is. Here is a corrected version:
The result follows from three standard results.
- Let $\tau(G)$ be the vertex covering number of $G$; then $\alpha(G)+\tau(G)=|V(G)|$.
- Let $\nu(G)$ be the matching number of $G$; then $\nu(G)+\beta'(G)=|V(G)|$ if $G$ has no isolated vertex.
- (König) If $G$ is bipartite, then $\tau(G)=\nu(G)$.
Thus, if $G$ is bipartite and has no isolated points,
$$\beta'(G)=|V(G)|-\nu(G)=|V(G)|-\tau(G)=\alpha(G)\;.$$
The first two are fairly easy to prove.
For the first one, suppose that $X$ is a vertex cover such that $|X|=\tau(G)$. Then $V(G)\setminus X$ is an independent set of vertices, so $\alpha(G)\ge|V(G)|-\tau(G)$, i.e., $$\alpha(G)+\tau(G)\ge|V(G)|\;.$$ On the other hand, if $X$ is an independent set of vertices, and $|X|=\alpha(X)$, then $V(G)\setminus X$ is a vertex cover, so $\tau(G)\le|V(G)|-\alpha(G)$, i.e., $$\alpha(G)+\tau(G)\le|V(G)|\;.$$
For the second, let $F$ be an edge cover such that $|F|=\beta'(G)$. From each component of the graph $\langle V(G),F\rangle$ choose one edge; the result is a matching $M$. The graph $\langle V(G),F\rangle$ has at least $|V(G)|-|F|$ components — $\langle V(G),\varnothing\rangle$ has $|V(G)|$ components, and adding an edge reduces the number of components by at most $1$ — so $$\nu(G)\ge|M|\ge|V(G)|-|F|=|V(G)|-\beta'(G)\;,$$ i.e., $$\nu(G)+\beta'(G)\ge|V(G)|\;.$$
On the other hand, let $M$ be a matching such that $|M|=\nu(G)$. Let $W=V(G)\setminus V(M)$; since $M$ is maximal, $W$ is an independent set of vertices. There are no isolated vertices, so for each $w\in W$ we may choose an edge $e_w$ incident at $w$; let $F=\{e_w:w\in W\}$. Then $F\cup E(M)$ is an edge cover of $G$, so
$$\begin{align*}
\beta'(G)&\le|F|+|E(M)|=|V(G)|-|V(M)|+|E(M)|\\
&=|V(G)|-2\nu(M)+\nu(M)=|V(G)|-\nu(M)\;,
\end{align*}$$
i.e.,
$$\nu(G)+\beta'(G)\le|V(G)|\;.$$
For König’s theorem see here (and many other places).
Best Answer
In as many words, they mean that the union of all maximal matchings contains all the edges incident on some $x\in X$.