Let $\{x_n\}$ and $\{y_n\}$ denote two bounded sequences. Prove that: $$ \lim \inf x_n + \lim \sup y_n \le \lim \sup(x_n + y_n) \\ $$
We know that both $x_n$ and $y_n$ are bounded hence is their sum: $$ m \le x_n + y_n < M $$
Using that fact we may choose a subsequence in order to satisfy the following: $$ \lim(x_{n_k} + y_{n_k}) = \lim\sup(x_n + y_n) \tag1 $$
Since $x_{n_k}$ is bounded (as far as $x_n$ is) lets choose a convergent subsequence with indices $n^\prime_k \ge n_k$ such that: $$ \exists \lim x_{n^\prime_k} $$
Now consider a sequence $y_{n^\prime_k}$ (note the index is $n^\prime_k$), since it is bounded we may choose a convergent subsequence from $y_{n^\prime_k}$ with indices $n^{\prime\prime}_k \ge n^\prime_k$ such that: $$ \exists\lim y_{n^{\prime\prime}_k} $$
Since $\{x_{n^{\prime\prime}_k}\}$ is a subsequence of $\{x_{n^\prime_k}\}$ it is convergent to the same limit. Also $\{y_{n^{\prime\prime}_k}\}$ is a subsequence of $\{y_{n^\prime_k}\}$ and we've chosen $\{y_{n^{\prime\prime}_k}\}$ to be convergent. Based on that and on $(1)$ we may write: $$ \lim(x_{n^{\prime\prime}_k} + y_{n^{\prime\prime}_k}) = \lim(x_{n_k} + y_{n_k}) = \lim\sup (x_n + y_n) \tag 2 $$
By definition of limsup and liminf: $$ \lim x_{n^{\prime\prime}_k} \ge \lim\inf x_n \\ \lim y_{n^{\prime\prime}_k} \le \lim\sup y_n $$
Or (multiply second inequality by $-1$): $$ \lim x_{n^{\prime\prime}_k} \ge \lim\inf x_n \\ -\lim y_{n^{\prime\prime}_k} \ge -\lim\sup y_n $$
Subtract the inequalities: $$ \lim x_{n^{\prime\prime}_k} + \lim y_{n^{\prime\prime}_k} \ge \lim\inf x_n + \lim\sup y_n \tag3 $$
Limit of sum is just a sum of limits so: $$ \lim(x_{n^{\prime\prime}_k} + y_{n^{\prime\prime}_k}) =\lim x_{n^{\prime\prime}_k} + \lim y_{n^{\prime\prime}_k} $$
So now using $(2)$ and $(3)$ we conclude that: $$ \lim \sup(x_n + y_n) \ge \lim\inf x_n + \lim\sup y_n $$
Is this argument enough to consider the proof complete?