From Baby Rudin chapter 3
For any two real sequences $\{a_n\}$, $\{b_n\}$, prove that $$\limsup_{n\to\infty} (a_n + b_n) \leq \limsup_{n\to\infty} a_n + \limsup_{n\to\infty} b_n$$ provided the sum on the right is not of the form $\infty - \infty$
Edit: Here is a second attempt using Rudin's definition via limits of sub sequences. I deleted my first attempt as it would've been confusing.
First, if $\limsup a_n$ or $\limsup b_n$ are equal to $\infty$ then $\limsup (a_n + b_n)$ will always be less than or equal to $\infty$
Now, assume $A, B, C$ are the finite $\limsup$ of $a_n, b_n,$ and $(a_n + b_n)$ respectively and that $C > A + B$. Then there exists some sub-sequence $c_{n_k} = a_{n_k} + b_{n_k}$ such that $c_{n_k} \to C$ as $k\to \infty$. Pick $\epsilon = C - (A+B) > 0$, then $\exists N$ such that $|c_{n_k} - C| < \epsilon$ when $k > N$. Therefore there exist infinitely many $c_{n_k} > A + B \implies (a_{n_k} - A) + (b_{n_k} - B) > 0$. So either $a_{n_k} - A >0$ or $b_{n_k} - B > 0$ for $k>N$ This yields a contradiction.
$a_j+b_j \leq \sup \{a_k:k\geq n\}+\sup \{b_k:k\geq n\}$ provided $j \geq n$. Take supremum over all $j \geq n$ to get $\sup \{a_j+b_j:j\geq n\} \leq \sup \{a_k:k\geq n\}+\sup \{b_k:k\geq n\}$. Tke lmit as $n \to \infty $ to get $C \leq A+B$.