I will write only about the first inequality $\limsup\limits_{n \rightarrow \infty} s_n + \liminf\limits_{n \rightarrow \infty} t_n \leq \limsup\limits_{n \rightarrow \infty} (s_n + t_n).$ (You wrote that you can prove the rest.)
By definition, if $\liminf t_n=t$ then for each $\varepsilon>0$ the inequality $t_n>t-\varepsilon$ holds for all but finitely many $n$'s. For such $n$'s we also have $s_n+t_n>s_n+t-\varepsilon$ and $\limsup(s_n+t_n) \ge \limsup (s_n+t-\varepsilon) = t-\varepsilon+ \limsup (s_n).$ (We have used monotonicity of $\limsup$ and that $\limsup (C+x_n)=C+\limsup x_n$ for any constant $C$.) Since the above inequality is true for each $\varepsilon>0$, we get that $\limsup(s_n+t_n) \ge t + \limsup (s_n)=\liminf t_n+\limsup s_n.$
EDIT: Note that the above proof does not work for $t=-\infty$ (it does not make sense to write $-\infty-\varepsilon$), but in this case the inequality is clear. (Of course, we have to omit indeterminate case $\infty-\infty$, i.e., in this case we assume that $\limsup s_n$ is finite.)
Or if you use $\liminf\limits_{n\to\infty} x_n= \lim\limits_{n\to\infty} \inf\limits_{k\ge n} x_k$ and $\limsup\limits_{n\to\infty} x_n= \lim\limits_{n\to\infty} \sup\limits_{k\ge n} x_k$ as the definition of limit inferior/superior then you can use $\sup_{k\ge n} (x_k+y_k) \ge \sup_{k\ge n} x_k + \inf_{k\ge n} y_k$ to get $\lim_{n\to\infty}\sup_{k\ge n} (x_k+y_k) \ge \lim_{n\to\infty}\sup_{k\ge n} x_k + \lim_{n\to\infty}\inf_{k\ge n} y_k\\ \limsup_{n\to\infty} (x_n+y_n) \ge \limsup_{n\to\infty} x_n + \liminf_{n\to\infty} y_n.$