Prove that $\limsup _{n\to \infty}(a_n+b_n)\leq \limsup _{n\to \infty}a_n + \limsup _{n\to \infty}b_n$ [duplicate]

Proposition: Let $\limsup\limits_{n\rightarrow \infty}{x_n}=x$ and let $x<\alpha$, then there is an integer $N$ such that $n\ge N$ implies $x_n<\alpha$.

Proof: Suppose there is an $\alpha>x$ such that for every $n\ge N$ implies $x_n\ge \alpha$ for infinitely many values of $n$. Let $A$ be the set of all such $x_n$, if $A$ has an upper bound then by least upper bound property of $\mathbb{R}$, $A$ has a supremum, $\gamma\ge\alpha>x$. Otherwise there is a subsequence $x_{n_k}$ such that $x_{n_k}\rightarrow +\infty$. Both cases contradict the definition of $\limsup. \qquad \square$

Let $\limsup\limits_{n\rightarrow \infty}{a_n}=a$ and $\limsup\limits_{n\rightarrow \infty}{b_n=b}$ as it is given that the sequences $a_n$ and $b_n$ are bounded, hence $a$ and $b$ are real.

For arbitrary $\varepsilon>0$ there exits an $N$ such that $a_n<a+\frac{\varepsilon}{2}$ and $b_n<b+\frac{\varepsilon}{2}$. Hence $$\begin{array}{rcl}a_n+b_n&<&a+b+\varepsilon\implies\\\limsup\limits_{n\rightarrow \infty}{(a_n+b_n)}&\le& a+b+\varepsilon \implies \\ \limsup\limits_{n\rightarrow \infty}{(a_n+b_n)}&\le&\limsup\limits_{n\rightarrow\infty}{a_n}+\limsup\limits_{n\rightarrow\infty}{b_n} \end{array}$$

Note: The following is also true but with a weaker hypothesis.

For any two real sequences $\{a_n\}$ and $\{b_n\}$ $$\limsup\limits_{n\rightarrow \infty}{(a_n+b_n)}\le \limsup\limits_{n\rightarrow \infty}{a_n} +\limsup\limits_{n\rightarrow \infty}{b_n}$$ provided that the right hand side is not of the form $\infty-\infty$. You can see here for a proof.

Notice that $$ \limsup a_n= \lim_{n\rightarrow\infty}\sup_{m\geq n}a_m, $$ Thus, you can use the linearity of the limit together with $$ \sup_{m\geq n}(a_m+b_m)\leq \sup_{m\geq n}a_m+\sup_{m\geq n}b_m. $$ Right?