Here is Prob. 41, Chap. 1, in the book Real Analysis by H.L. Royden and P.M. Fitzpatrick, 4th edition:
Show that $\lim\inf a_n \leq \lim\sup a_n$.
Here $\left( a_n \right)$ is of course a sequence of real numbers. Moreover, $$ \lim\inf a_n := \lim_{n \to \infty} \inf \left\{ a_k \, | \, k \in \mathbb{N}, k \geq n \right\} = \sup_{n \in \mathbb{N}} \inf_{k \in \mathbb{N}, k \geq n} a_k, $$ and $$ \lim\sup a_n := \lim_{n \to \infty} \sup \left\{ a_k \, |\, k \in \mathbb{N}, k \geq n \right\} = \inf_{n \in \mathbb{N}} \sup_{k\in \mathbb{N}, k \geq n} a_k, $$ where the $\inf$ and the $\sup$ in each case is taken over the extended real number system.
My Attempt:
For any natural numbers $m$ and $n$ such that $m < n$, we find that $$ \inf_{k \in \mathbb{N}, k \geq m} a_k \leq \inf_{k \in \mathbb{N}, k \geq n} a_k \leq \sup_{k \in \mathbb{N}, k \geq n} a_k, $$ and aslo $$ \inf_{k \in \mathbb{N}, k \geq n} a_k \leq \sup_{k \in \mathbb{N}, k \geq n} a_k \leq \sup_{k \in \mathbb{N}, k \geq m} a_k. $$
Thus for any natural numbers $n$ and $m$, we have $$ \inf_{k \in \mathbb{N}, k \geq m} a_k \leq \sup_{k \in \mathbb{N}, k \geq n} a_k. $$ So for any natural number $m$, we have $$ \inf_{k \in \mathbb{N}, k \geq m} a_k \leq \inf_{n \in \mathbb{N} } \sup_{k \in \mathbb{N}, k \geq n} a_k = \lim \sup a_n, $$ which in turn implies that $$ \lim \inf a_n = \sup_{m \in \mathbb{N} } \inf_{k \in \mathbb{N}, k \geq m} a_k \leq \lim \sup a_n, $$ as required.
Is this proof correct and clear enough in each and every detail? Or, are there any issues with it?