ORIGINAL (FLAWED) PROOF
Without loss of generality, let $G$ be connected. Let $e$ be a cut-edge with endpoints $u$ and $v$. Since $e$ is a cut-edge, its removal would separate $G$ into two components $H_1$ and $H_2$. This means that every path from a vertex in $H_1$ to a vertex in $H_2$ passes through $e$, and so every such path passes through both $u$ and $v$. Therefore, both $u$ and $v$ are cut-vertices.
ADDENDUM
As Joseph mentioned in the comments, a pendant edge is always a cut-edge, while a leaf cannot be a cut-vertex. Thus, the only way that both $u$ and $v$ fail to be cut-vertices is thus the case when $G$ is a single edge.
REVISED PROOF
Without loss of generality, let $G$ be connected. Let $e$ be a cut-edge with endpoints $u$ and $v$.
Note that if a vertex is a leaf, it cannot be a cut-vertex, since its removal does not increase the number of components of $G$. Thus, if $e$ is the only edge of $G$, then both $u$ and $v$ are leaves, and so $G$ possesses no cut-vertex.
Assume now that $G$ contains another edge $f$ distinct from $e$. Since $e$ is a cut-edge, its removal would separate $G$ into two components $H_u$ and $H_v$. This means that every path from a vertex in $H_u$ to a vertex in $H_u$ passes through $e$, and so every such path contains both $u$ and $v$. Now, one of $H_u$ or $H_v$ contains a vertex other than $u$ or $v$ (respectively), as $f$ belongs to one of these two components. Therefore, one of $u$ or $v$ is a cut-vertex. Furthermore, if neither $u$ nor $v$ is a leaf, then both $u$ and $v$ are cut-vertices.
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
Suppose that there is a cut vertex $v$, and let $X, Y$ be two sets of vertices separated by $v$.
Since $G$ is $\alpha$-critical, you should be able to prove that there is a maximum independent set $I$ that contains $v$. Let $\alpha_X$ (respectively $\alpha_Y$) be the number of vertices of $X$ (respectively $Y$) included in $I$. Thus $\alpha(G) = \alpha_X + \alpha_Y + 1$.
Let $x \in X$ be a neighbor of $v$ in $X$. By removing the edge $vx$, by the $\alpha$-critical property we get an independent set $I'$ of size $I + 1$ that includes both $v$ and $x$. This is only possible if $I'$ includes $\alpha_X + 1$ vertices of $X$ (why?), and hence $X$ admits an independent set $I_X$ of size $\alpha_X + 1$. By the same argument, $Y$ admits an independent set $I_Y$ of size $\alpha_Y + 1$.
But then, $I_X \cup I_Y$ is an independent set of size $\alpha_X + \alpha_Y + 2$, which is greater than $\alpha(G)$, a contradiction.