Using Cantor's lemma in a proof of convergence

It is a valid proof - at least with the addendum of the argument JonathanZ has pointed out is missing: that the interval sizes go to $0$.

Your notation leaves something to be desired. Sets have infimums and supremums, not individual elements. You could have defined $A_k = \{a_i \mid i \ge k\}$ and then called them $\inf A_k$ and $\sup A_k$, or just called them $\inf \{a_i \mid i \ge k\}$ or even $\inf_{i \ge k} a_i$. Either one would be acceptable.

You also don't need Cantor's lemma - you've already made use of the completeness of $\Bbb R$ in claiming the existance of $m_k = \inf \{a_i \mid i \ge k\}$ and $M_k = \sup \{a_i \mid i \ge k\}$.

Since for all $i \ge k, m_k \le m_i \le a_i \le M_i \le M_k$, the sequence $\{m_i\}_{i=1}^\infty$ is increasing and bounded above by all $M_k$, while $\{M_i\}_{i=1}^\infty$ is decreasing an are bounded below by all $m_k$. Therefore $$\lim_{k \to \infty} m_k = \sup \{m_i\}_{i=1}^\infty = \liminf_k a_k \le \limsup_k a_k = \inf \{M_i\}_{i=1}^\infty = \lim_{k\to\infty} M_k$$

Therefore, your argument can be simplified to:

Since $\{a_k\}_{k=1}^\infty$ is bounded, $\liminf_k a_k$ and $\limsup_k a_k$ exist and since $\sup_{i \ge k} a_i - \inf_{i \ge k} a_i\to 0$ as $k \to \infty$, we have $\liminf_k a_k = \limsup_k a_k$. Therefore $\lim_k a_k$ converges to the common value.

Add actual arguments showing that $a_k$ is bounded and that $\sup_{i \ge k} a_i - \inf_{i \ge k} a_i\to 0$, and you are done.