Late answer, but for future reference (so that anyone reading this thread won't spend as much time as I did trying to prove the conjecture) the conjecture is false, with the following simple graph as a counterexample. Let $G$ be the following graph

One can see that $\alpha(G) = 4$. It is also easy to see that if we remove any of its edges we obtain a graph with independence number 5, whence it is $\alpha$-critical. The vertex set $S = \{1,4,7,9\}$ (or, alternatively, $S=\{2,5,8,9\}$) is a maximum independent set with the property that $H = G \setminus S$ (the graph formed by removing the vertices in $S$, and all edges incident to them, from $G$) is just the graph consisting of three $K_2$-components. Thus $\alpha(H) = 3$ and therefore the maximum size of any independent set in $G$ disjoint from $S$ is three.

Another reason for reopening this thread is that I would be interested to know if the following weakening of the conjecture could be true:

**Weak conj.** for every $\alpha$-critical graph $G$ without isolated vertices, there exists a maximum independent set $S$ in $G$ and a maximum independent set $S′$ in $G$ that is disjoint from $S$.

I somehow get the feeling that this is unlikely to be true, however I am unable to find a counterexample. The graph above, that provides a counterexample to the stronger conjecture, does indeed satisfy this weaker version, for example by taking the maximum independent sets $S = \{1,2,4,9\}$ and $S' = \{5,6,7,8\}$. There is one more (connected) graph with a counterexample to the stronger conjecture that I know of;

with $S = \{1,4,7,9\}$.

However, also this graph satisfies the weaker version of the conjecture (take e.g. $S = \{1,3,4,9\}$ and $S' = \{0,6,7,8\}$).

I know of no more counterexamples to the stronger conjecture and I believe that the only graphs that provide counterexamples with less than $11$ vertices are the two graphs mentioned above. Thus I believe that any counterexample to the weaker conjecture would have to have at least $11$ vertices.