I got this proof from my professor during office hours:
Let $C$ be a minimum cover, and let
$$C_1 = |C\cap A|$$
$$C_2 = |C\cap B|\;.$$
So since the size of a maximum matching is equal to $|C|$, suppose by contradiction that
$$|C| < \min(n, (q - \delta^2)/(n- \delta))\;.$$
$$C_1 + C_2 < n$$
$$C_2 \geq \delta, C_1 \geq \delta\;.$$
As there can't be any edges between $A\setminus C$ and $B\setminus C$, we can bound the number $q$ of edges by adding up all possible edges between $A\cap C$ and $B\setminus C$, between $A\setminus C$ and $B\cap C$ and between $A\cap C$ and $B\cap C$:
$$ q \leq C_1(n - C_2) + C_2(n - C_1) + C_1C_2\;.$$
He said this next step is the crucial move with a little wishful thinking. Basically he said he wants to turn
$$ q \leq C_1(n - C_2) + C_2(n - C_1) + C_1C_2$$
$$ q \leq C_1(n - \delta) + C_2(n - \delta) + \delta^2\;.$$
But we can't just replace $C_1C_2$ by $\delta^2$ since $\delta^2\leq C_1C_2$, so he used the fact that
$$ (C_1 - \delta)(C_2 - \delta) \geq 0$$
since $C_2 \geq \delta, C_1 \geq \delta$. So
C_1(n - C_2) + C_2(n - C_1) + C_1C_2
(C_1 + C_2)n - (C_1 + C_2)\delta + \delta^2 - (C_1C_2 - C_1\delta - C_2\delta + \delta^2)
|C| (n - \delta) + \delta^2 - (C_1 - \delta)(C_2 - \delta)
|C| (n - \delta) + \delta^2\;,
$$|C| \geq (q-\delta^2) / (n - \delta)\;,$$
which contradicts our assumption that $|C| < \min(n, (q - \delta^2)/(n- \delta))$.
(He said this proof is not very constructive but somehow it works)
As for the example,
consider a graph where the first two vertexes in $A$ and $B$ has degree $8$. The rest of the six vertexes from $A$ join the 2 vertexes in $B$ and vice versa (so the minimum degree for each of the six vertexes is $2$). This graph has maximum match of size $4$ because the initial four vertexes from $A$ and $B$ are the minimum cover.
this is roughly the graph (if you don't mind the bad drawing):
So $q$ = $2 \times 6 + 2 \times 8 = 28$ edges.
So applying the bound $(28 - 4) / (8 - 2) = 4$ which holds since $|M| = 4$, but the tighter bound is $(28 - 8) / (8 - 4) = 5$.