Let $(x_n)$ and $(y_n)$ be bounded sequences. Then there exist real numbers $M$ and $N$ such that $x_n \leq M$ and $y_n \leq N$ for all $n$. Let $L = \limsup(x_n)$ and $K = \limsup(y_n)$. Then for any $\epsilon > 0$, there exists an integer $N_1$ such that for all $n \geq N_1$, we have $$ L - \epsilon < x_n \leq L + \epsilon. $$ Similarly, there exists an integer $N_2$ such that for all $n \geq N_2$, we have $$ K - \epsilon < y_n \leq K + \epsilon. $$ Let $N_3 = max(N_1,N_2)$. Then for all $n\ge N_3$, we have $$ (L+K)-2\epsilon < x_n+y_n\le (L+K)+2\epsilon. $$
Taking the limit superior of both sides as n goes to infinity gives us: $$ (L+K)-2\epsilon\le limsup(x_n+y_n)\le (L+K)+2\epsilon. $$
Since $\epsilon>0$ was arbitrary, it follows that: $$limsup(x_n+y_n)\le limsup(x_n)+limsup(y_n).$$
Can anyone see if my proof is alright?
You can show that forall $n \in \mathbb{N}$, $$ \sup_{m \ge n} (x_m + y_m) \le \sup_{m \ge n} x_m + \sup_{m \ge n} y_m$$ by simply showing the $\sup_{m \ge n} x_m + \sup_{m \ge n} y_m$ is an upper bound of the set $\{ x_m + y_m | m \ge n \}$. And then just take limit on both sides, you know it exists in $\mathbb{R}$ because the sequences are given to be bounded, and the result follows by linearity of limit over sums.