Problem:
For graph $F$, let $\alpha(F)$ denote the maximal size of $F$'s independent vertices set.
$G$ is a connected simple graph such that $\alpha(G)=a$, and for any edge $e$ of $G$, $\alpha(G-e)=a+1$, where $G-e=(V(G), E(G)-e)$. Show that $a\leqslant \frac{1}{2}|V(G)|$.
Source: own.
Attempt:
I thought of induction on $a$, but met with some obstruction (can't transfer to the lower situation). Then I thought of induction on a stronger conclusion, like:
$G$ is a connected simple graph such that for any vertex $v\in G$, there is an independent set of $G$ of size $a$ that contains $v$, and $a$ can't be larger. For any edge $e$ of $G$, there is an independent set of $G-e$ of size $a+1$ that contains both vertices of $e$. Then prove that $a\leqslant \frac{1}{2}|V(G)|$.
But it becomes too puzzling and I can't ensure if it's a correct one. Please help. (I hope there is a solution without induction.)