I've been self studying real analysis and I'm trying to prove $\limsup_{n\to\infty}(a_n+b_n)\leq\limsup_{n\to\infty}a_n+\limsup_{n\to\infty}b_n$ using the definition $\limsup_{n\to\infty}a_n=\lim_{n\to\infty}\left(\sup_{k\ge n}a_n\right)$. Here's my attempt:
The strategy is to show $\sup_{k\ge n}(a_k+b_k)\leq\sup_{k\ge n}a_n+\sup_{k\ge n}b_n$, then use the order limit theorem to establish the inequality in the limiting process.
First, we see that $\sup_{k\ge n}a_n\ge a_k$ and $\sup_{k\ge n}b_n\ge b_k$ for all $k\ge n$, which implies $\sup_{k\ge n}a_n+\sup_{k\ge n}b_n\ge a_k+b_k$ for all $k\ge n.$ In particular, this shows $\sup_{k\ge n}a_n+\sup_{k\ge n}b_n$ is an upper bound of the set $\{a_k+b_k:k\ge n\}$, so we have that $\sup_{k\ge n}(a_k+b_k)\leq\sup_{k\ge n}a_n+\sup_{k\ge n}b_n$ because the left hand side is the least upper bound of the set. Now take the limit on both sides and we have $$\lim_{n\to\infty}\left(\sup_{k\ge n}(a_k+b_k)\right)\leq\lim_{n\to\infty}\left(\sup_{k\ge n}a_n\right)+\lim_{n\to\infty}\left(\sup_{k\ge n}b_n\right)$$ or $$\limsup_{n\to\infty}(a_n+b_n)\leq\limsup_{n\to\infty}a_n+\limsup_{n\to\infty}b_n.$$
Is my proof valid?
