Why $M_1 \subset M_2 \not \Rightarrow N_{M_1} (\lambda) \leq N_{M_2} (\lambda)$ for eigenvalue problem? (EDIT)

In view of Terry's comment on Michael's answer, it's perhaps worth pointing out that this monotonicity also fails if both domains are required to be convex. We can take $M_2=[0,L]^2$ as a square of side length $L$. Then $\lambda_2(M_2)=\pi^2/L^2$ (possible eigenfunction $u=\cos \pi x/L$). If we now take $M_1\subseteq M_2$ as a thin rectangle close to the diagonal, then we obtain approximately the one-dimensional Neumann eigenvalue of an interval of that length: $$ \lambda_2(M_1)\simeq \pi^2/(\sqrt{2}L)^2<\lambda_2(M_2) $$

Take $M_1$ to be a region consisting of two "blobs" connected by a narrow channel of length 1 and width $\epsilon$. Now choose $u_1$ and $u_2$ such that each is 1 in one of the blobs, 0 in the other, and linearly interpolated in the narrow channel. It follows that both $\int |\nabla u_1|^2$ and $\int |\nabla u_2|^2$ are order $\epsilon$. Consequently $N_{M_1}(\lambda)\ge 2$ if $\lambda>>\epsilon$. Now take $M_2$ to be a ball containing $M_1$. In $M_2$, the second eigenvalue of the Neumann problem is of order 1.