= Solution
The intended deletion theorem starts from the full action set $A=M\cup N$. Under that interpretation, each player's surviving action set remains nonempty: with only one action, no <mixed strategy> can weakly improve it and improve it strictly somewhere. Let the final reduced <bimatrix game> have a <Nash equilibrium>, which exists by <Nash's theorem>.
Restore the deleted actions in reverse order. Consider a row action $i$ when it is restored. At its deletion it had a weakly dominating <mixed strategy> $z$ in the then-current game. If $z_i>0$, remove that self-weight and renormalize: the strict-improvement clause guarantees $z_i<1$, and dividing the payoff comparison by $1-z_i$ gives a weakly dominating strategy supported on the other then-current row actions. Those other actions have already been restored in the reverse process. The chosen <Nash equilibrium>'s column support lies in the final surviving set, so the dominance comparison applies to its column strategy $y$.
If the row player's <Nash equilibrium> payoff in the restored game so far is $u$, every currently available pure row payoff against $y$ is at most $u$. The dominating mixture therefore has payoff at most $u$, and the restored action has payoff no greater than that mixture. It cannot improve the payoff. Column deviations are unchanged by adding a row action, since the actual <Nash equilibrium> strategy remains fixed. The analogous argument works when restoring a column action. Induction restores the full game while preserving the final reduced <Nash equilibrium> and its support.
Thus \b[iterated deletion of weakly dominated actions preserves at least one <Nash equilibrium> supported entirely on surviving actions]. It need not preserve every <Nash equilibrium> or give an order-independent reduced game.
The literal printed use of an arbitrary initial $A$ needs qualification. For example, take row payoffs $\left(\begin{smallmatrix}0&3\\1&0\end{smallmatrix}\right)$ and column payoffs $\left(\begin{smallmatrix}0&1\\0&1\end{smallmatrix}\right)$. On the restricted set containing both rows but only column $1$, row $1$ is strictly dominated by row $2$. Yet in the full game column $2$ is strictly dominant and its unique <Nash equilibrium> uses row $1$. Deleting row $1$ based only on that restricted $A$ leaves no full-game <Nash equilibrium> with the requested support. The proven <equilibrium preservation under iterated weak dominance> therefore requires \b[initial $A=M\cup N$], or concludes only about the initially restricted game.
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