Prove two inequalities about limit inferior and limit superior I wish to prove the following two inequalities: 

Suppose $X$ is a subset in $\Bbb R$, and functions $f$ and $g$: $X\to \Bbb R$, and $x_{0}\in X$ is a limit point. Then: $$\lim\sup_{x\to x_0}(f(x)+g(x))\le \lim\sup_{x\to x_0}(f(x))+\lim\sup_{x\to x_0}(g(x))$$  and,$$\lim\inf_{x\to x_0}(f(x))+\lim\inf _{x\to x_0}(g(x))\le \lim\inf_{x\to x_0}(f(x)+g(x)).$$

I did try to use proof by contradiciton, but I just got the desired results. Any help on them, please. 
 A: Contradiction is not recommended, as there is a natural direct approach.
1- limsup: recall that 
$$
\limsup_{x\rightarrow x_0}h(x)=\inf_{\epsilon>0}\sup_{0<|x-x_0|<\epsilon}h(x)=\lim_{\epsilon>0}\sup_{0<|x-x_0|<\epsilon}h(x)
$$
where the rhs is the limit of a nonincreasing function of $\epsilon$. Note that the condition $x\in X$ should be added everywhere we take the a sup above. Let's say it is here implicitly to alleviate the notations.
The only thing you need to obtain the desired inequality is essentially the fact that for two nonempty subsets $A,B$ of $\mathbb{R}$
$$
\sup (A+B)\leq \sup A+\sup B.
$$
I prove it at the end below. Now fix $\epsilon>0$. The latter provides the second inequality in the following
$$
\limsup_{x\rightarrow x_0}f(x)+g(x)\leq \sup_{0<|x-x_0|<\epsilon}f(x)+g(x)\leq \sup_{0<|x-x_0|<\epsilon}f(x)+\sup_{0<|x-x_0|<\epsilon}g(x),
$$
while the first inequality is simply due to the fact that the limsup is the inf, hence a lower bound, of the $\epsilon$ suprema. 
The desired inequality follows by letting $\epsilon$ tend to $0$ in the rhs.
2- liminf: you can use a similar argument with $\inf (A+B)\geq \inf A+\inf B$. Or you can simply use the following standard trick
$$
\liminf_{x\rightarrow x_0}h(x)=-\limsup_{x\rightarrow x_0}-h(x)
$$
which follows from $\inf S=-\sup (-S)$ (good exercise on sup/inf upper/lower bound manipulation). It only remains to apply the limsup inequality to $-f$ and $-g$.
Sup inequality proof: it is trivial if one of the two sets $A,B$ is not bounded above. So assume both are bounded above. Now recall that $\sup S$ is the least upper bound of the set $S$, when it is bounded above. In particular, it is an upper bound of $S$. For every $a\in A$, we have $a\leq \sup A$, and for every $b\in B$, we get $b\leq \sup B$. So for every $x=a+b\in A+B$, we have $x\leq \sup A+\sup B$. Thus $\sup A+\sup B$ is an upper bound of $A+B$. Hence it is not smaller than the least upper bound of the latter, namely $\sup(A+B)$. QED.
