When you are asked to prove something trivial and obvious... well, do the definitions. I think your prove relies on "common sense" and makes jumps that are simply too big, to the point you are mostly just restating the statements.
I'd do: If $m \in M_1$ then $m \in M_2$ as $M_1 \subset M_2$. So $\inf M_2 \le m$ by definition of infinum. So $\inf M_2$ is a lower bound of $M_1$. But $\inf M_1$ is the greatest lower bound of $M_1$ so $\inf M_2 \le \inf M_1$.
By identical argument: $\sup M_2 \le \sup M_1$. That is; if $m \in M_1$ then $m \in M_2$ and therefore $\sup M_2 \ge m$ and is an upper bound of $M_1$, but as $\sup M_1$ is the least upper bound $\sup M_2 \ge \sup M_1$.
By definition of supremum and infimum for any $m \in M_1$ then $\inf M_1 \le m \le \sup M_1$ so $\inf M_1 \le \sup M_1$. (Unless $M_1$ is empty, in which case neither $\inf M_1$ nor $\sup M_1$ exist.) (I suppose at the very beginning of the proof I could/should have made a comment that $M_1$ and $M_2$ are both presumed to be non-empty and bounded above and below.).
Hence ... result.
That's basically your proof but with the jumps explicitly explained. (Perhaps painfully so.)
And, to be fair, maybe I'm being to0 glib in claiming "$\inf M_1$ is greatest lower bound".
More detail: If $\inf M_1 < \inf M_2$ then $\inf M_2$ is not a lower bound of $M_1$ which we just showed it is, so $\inf M_2 \le \inf M_1$.
That's kind of obvious but as I criticized you for not proving the obvious, it's only fair that I apply my own standards...