Here is my attempt. Any indication will be appreciated.
$$y_n\rightharpoonup0 \Rightarrow \,\forall k:\,\,<y_n,e_k>\xrightarrow{n\rightarrow\infty}0$$ (Since $e_k \in {l_p}^* = l_q $ , $q$ is conjugate index of $p$)
Also, since $c_{00} = \{\vec{a}: a(n)\in F\text{, number of nonzero element is finite}\}$ dense in $(l_p,\|\cdot\|_{l_p})$, there is $(x_m) \in c_{00}$ such that $\,x_m \rightarrow x \,\,\text{ in } \,\,(l_p,\|\cdot\|_{l_p})$.
Now, you can easily verify (like in your comment) followings. (for all m)
$$
\underset{n\to\infty}{\lim\sup}\,{\|x_m+y_n\|}
^p = {\|x_m\|}^p+\underset{n\to\infty}{\lim\sup}\,{\|y_n\|}^p\,\,\text{ - (1)}
$$
Moreover, by operating ' $\underset{m\to\infty}{\lim\sup} \,\, \underset{n\to\infty}{\lim\sup}$' on following inequality you can get $\underset{m\to\infty}{\lim\sup} \,\, \underset{n\to\infty}{\lim\sup}{\,\|x_m+y_n\|} = \underset{n\to\infty}{\lim\sup}{\|x+y_n\|}$ - (2).
$$
-{\|x_m - x\|}\leq{\|x_m+y_n\|}-{\|x+y_n\|}\leq{\|x_m - x\|}
$$
Then from (2), $\underset{m\to\infty}{\lim\sup} \,\, \underset{n\to\infty}{\lim\sup}{\,\|x_m+y_n\|}^p = \underset{n\to\infty}{\lim\sup}{\|x+y_n\|}^p$ - (3) is also true.
Finally, using (3) on $\underset{m\to\infty}{\lim\sup}$ (1) will give your equality.
Addendum
First, there is $N$ such that $k>N \Rightarrow x_m (k) = 0$.
for $\epsilon'>0$, there is $N'$ such that $$n>N' \Rightarrow |y_n(k)| < \epsilon = \frac{1}{2} \min[\epsilon'/N^{1/p} ,pN(2\|x_m\|_{sup})^{1-p} ,\|x_m\|_{sup}] \, , \forall k\leq N$$
then, for $n>N'$ (N is fixed),
$$
\Bigg|\|x_m+y_n\|^p - (\|x_m\|^p + \|y_n\|^p) \Bigg| \\= \Bigg|\sum_{k\leq N} |x_m(k) + y_n(k)|^p + \sum_{k > N} | y_n(k)|^p - \sum_{} |x_m(k)|^p - \sum | y_n(k)|^p \Bigg| \\= \Bigg|\sum_{k\leq N} (|x_m(k) + y_n(k)|^p - |x_m(k)|^p) - \sum_{k \leq N} | y_n(k)|^p \Bigg|\\ \leq pN\epsilon (\|x_m\|_{sup} + \epsilon)^{p-1} + N\epsilon^p \text{ (MVT used for first part) }\\<\epsilon'
$$
Thus
$$
\|x_m+y_n\|^p < \|x_m\|^p + \|y_n\|^p + \epsilon' \quad\forall n>N'\\ $$
by '$\underset{n\to\infty}{\lim\sup}$' on each side,
$$
\underset{n\to\infty}{\lim\sup}\,\|x_m+y_n\|^p < \|x_m\|^p + \underset{n\to\infty}{\lim\sup}\,\|y_n\|^p + \epsilon'.
$$
Since $\epsilon'$ is arbitrary, $\underset{n\to\infty}{\lim\sup}\,\|x_m+y_n\|^p \leq \|x_m\|^p + \underset{n\to\infty}{\lim\sup}\,\|y_n\|^p$.
Another inequality ($\geq$) is almost same.
Want: $({\lim\sup}\,{a_n})^p = {\lim\sup}\,{a_n}^p$, when $p>0$ and $a_n \geq 0\\$.
First, from definition of limsup,
$$
({\lim\sup}\,{a_n})^p = (\underset{n}{\inf}\,[{\sup}\,\{x_k \,:\, k\geq n \}])^p
$$
Since $x \mapsto x^p$ is continuous and monotonically increasing function on $x\geq0$,
(Actually, there is the case for ${\sup}\,\{a_k \,:\, k\geq n \} = \infty$ , but it is easier to show $\infty = \infty$ of WANT.)
$$
=\underset{n}{\inf}\,[{\sup}\,\{a_k \,:\, k\geq n \}]^p
$$
Again, by same reason,
$$
=\underset{n}{\inf}\,[{\sup}\,\{a_k^p \,:\, k\geq n \}]
$$
Indeed,
$$
={\lim\sup}\,{a_n}^p \quad\text{(By definition)}
$$
All you have to show now is (and same for $\inf$)
$$ \sup f(a_n) = f [ \sup a_n ]$$
where $f$: continuous and mono. increasing.
For convenience, let $\alpha = \sup a_n$.
(case of $\alpha = \infty$ needs slightly different argument, but not the case of ours.)
since $a_n \leq \alpha$ for all $n$, $f(a_n) \leq f(\alpha)$ , thus $f(\alpha)$ is upper bound for $\{f(a_n)\}$ .
Thus $\boldsymbol{f(\alpha) \geq \sup f(a_n)}$.
For another inequality, enough to show $f(\alpha) < \sup f(a_n) + \epsilon$, for arbitrary $\epsilon >0$.
Since $f$ is continuous at $x = \alpha$, there is some $\delta>0$ such that $|\alpha - x|<\delta$ implies $|f(\alpha) - f(x) | <\epsilon$ .
Also, since $\alpha = \sup a_n$, there is $m\in\mathbb{N}$ such that $a_m + \delta > \alpha$. Thus, $|f(\alpha) - f(a_m) | <\epsilon$.
i.e.
$$
f(\alpha)<\epsilon + f(a_m)\, ,\,\, \boldsymbol{f(\alpha)<\epsilon + \sup f(a_n)}.
$$
Best Answer
Take the sequence $(x_n)$ in $l^\infty$ with $x_n=(0,\dots,0,1,1,\dots)$ (the first $n$ components are $0$). It is bounded and componentwise convergent to $0$. Thus if it is weaklyconvergent the weak limit would be $(0)$. Take a Banach limit $m \in (l^\infty)^\ast$. Then $m(x_n)=1$ $(n \in \mathbb{N})$. Thus $(x_n)$ is not weakly convergent.