I found an answer by my own.
We want to prove that the statement
$$
\exists C>0\;\;:||f+g||_{1,\infty}\le||f||_{1,\infty}+C||g||_{1,\infty}\;\;\;\forall f,g\in L^{1,\infty}(\Bbb R^n)
$$
is false (we should in fact expect this: looking at PART(A), we see that $C(\varepsilon)>0$ gets bigger and bigger as $\varepsilon$ comes closer to $0$, ). Thus we must prove that
$$
\forall C>0\;\;\exists f_C,\;g_C\in L^{1,\infty}(\Bbb R^n)\;\;:\;\; ||f_C+g_C||_{1,\infty}>||f_C||_{1,\infty}+C||g_C||_{1,\infty}\;\;.
$$
I took $n=1$ and I searched for two sequences of functions $\{f_k\}_k,\{g_k\}_k\subseteq L^{1,\infty}(\Bbb R)$ such that
\begin{equation}\label{4}
||f_k+g_k||_{1,\infty}>||f_k||_{1,\infty}+C||g_k||_{1,\infty}
\end{equation}
be true for some $k=k(C)$; in this way I would have finished.
Observe now that the above inequality is equivalent to
$$
\frac{||f_k+g_k||_{1,\infty}}{||g_k||_{1,\infty}}>\frac{||f_k||_{1,\infty}}{||g_k||_{1,\infty}}+C
$$
(we search for a $\{g_k\}_k$ such that $||g_k||_{1,\infty}\neq0\;\;\;\forall k$; otherwise if $||g_k||_{1,\infty}=0$ for a finite number of $k$'s or for infinitely many $k$'s but not definitively, we just drop these $g_k$ from the sequence; if otherwise $g_k=0$ a.e. definitively, then the above inequality becomes $||f_k||_{1,\infty}>||f_k||_{1,\infty}\;\;k\ge\bar k$, which is false).
Thus finding $\{f_k\}_k,\{g_k\}_k\subseteq L^{1,\infty}(\Bbb R)$ such that
\begin{equation}\label{5}
\left\{
\begin{array}{ll}
\limsup_k\frac{||f_k+g_k||_{1,\infty}}{||g_k||_{1,\infty}}=+\infty\\
\limsup_k\frac{||f_k||_{1,\infty}}{||g_k||_{1,\infty}}<+\infty\\
\end{array}
\right.
\end{equation}
would allow us to conclude.
\newline
\newline
I began to try with different functions on different domains, until the role of every component I was using became clear; then I could make an heuristic argument which finally gave two possible candidates for the desired sequences.
So let's $n=1$ and consider
$$
f_k(x):=-ke^x\chi_{]0,\sqrt k[}(x),\;\;\;\;\;\;\;g_k(x):=ke^x\chi_{]0,k[}\;\;,\;\;k\ge1
$$
from which
$$
f_k(x)+g_k(x)=ke^x\chi_{[\sqrt k,k[}(x)\;\;.
$$
Observe now, that, for a measurable function $f:\Bbb R^n\to\Bbb R$ we have that
$$
||f||_{1,\infty}:=\inf\left\{D>0\;:\;\lambda_f(\alpha)\le\frac D{\alpha}\;\forall\alpha>0\right\}=\sup_{\alpha>0}\{\alpha\lambda_f(\alpha)\}\;.
$$
We will use this last characterization of the $1$-weak norm in the following.
Now:
\begin{align*}
\lambda_{f_k}(\alpha)
&=
\left\{
\begin{array}{lll}
0\\
\sqrt k-\log\left(\frac{\alpha}{k}\right)\\
\sqrt k
\end{array}
\right.
\left.
\begin{array}{ccc}
\alpha\ge ke^{\sqrt k}\\
k\le\alpha<ke^{\sqrt k}\\
0<\alpha<k
\end{array}
\right.\\
&=\left(\sqrt k-\log\left(\frac{\alpha}{k}\right)\right)\chi_{[k,ke^{\sqrt k}[}(\alpha)+\sqrt k\chi_{]0,k[}(\alpha)
\end{align*}
from which we get
$$
||f_k||_{1,\infty}=\sup_{\alpha>0}\left\{\alpha\left(\sqrt k-\log\left(\frac{\alpha}{k}\right)\right)\chi_{[k,ke^{\sqrt k}[}(\alpha)+\alpha\sqrt k\chi_{]0,k[}(\alpha)\right\}=k\sqrt k
$$
in fact if we set $h(\alpha):=\alpha\left(\sqrt k-\log\left(\frac{\alpha}{k}\right)\right)$ we have $h'(\alpha)=\sqrt k-\log\left(\frac{\alpha}{k}\right)-k$ which is negative when $\alpha\in [k,ke^{\sqrt k}[$, thus here $h$ decreases, thus the $\sup$ above follows.
Let's now look at $g_k$:
\begin{align*}
\lambda_{g_k}(\alpha)
&=
\left\{
\begin{array}{lll}
0\\
k-\log\left(\frac{\alpha}{k}\right)\\
k
\end{array}
\right.
\left.
\begin{array}{ccc}
\alpha\ge ke^{k}\\
k\le\alpha<ke^{k}\\
0<\alpha<k
\end{array}
\right.\\
&=\left(k-\log\left(\frac{\alpha}{k}\right)\right)\chi_{[k,ke^{k}[}(\alpha)+k\chi_{]0,k[}(\alpha)
\end{align*}
from which we get
$$
||g_k||_{1,\infty}=\sup_{\alpha>0}\left\{\alpha\left(k-\log\left(\frac{\alpha}{k}\right)\right)\chi_{[k,ke^{k}[}(\alpha)+\alpha k\chi_{]0,k[}(\alpha)\right\}=k^2
$$
because, with the same argument as above, $h(\alpha):=\alpha\left(k-\log\left(\frac{\alpha}{k}\right)\right)$ is decreasing in $[k,ke^k[$.\
Finally let's work with the sum $f_k+g_k$:
\begin{align*}
\lambda_{f_k+g_k}(\alpha)
&=
\left\{
\begin{array}{lll}
0\\
k-\log\left(\frac{\alpha}{k}\right)\\
k-\sqrt k
\end{array}
\right.
\left.
\begin{array}{ccc}
\alpha\ge ke^{k}\\
ke^{\sqrt k}\le\alpha<ke^{k}\\
0<\alpha<ke^{\sqrt k}
\end{array}
\right.\\
&=\left(k-\log\left(\frac{\alpha}{k}\right)\right)\chi_{[ke^{\sqrt k},ke^{k}[}(\alpha)+(k-\sqrt k)\chi_{]0,ke^{\sqrt k}[}(\alpha)
\end{align*}
from which we get
$$
||f_k+g_k||_{1,\infty}=\sup_{\alpha>0}\left\{\alpha\left(k-\log\left(\frac{\alpha}{k}\right)\right)\chi_{[ke^{\sqrt k},ke^{k}[}(\alpha)+\alpha(k-\sqrt k)\chi_{]0,ke^{\sqrt k}[}(\alpha)\right\}=ke^{\sqrt k}(k-\sqrt k)
$$
again observing that $h(\alpha):=\alpha\left(k-\log\left(\frac{\alpha}{k}\right)\right)$ is decreasing in $[ke^{\sqrt k},ke^k[$.
Thus, such $f_k$'s and $g_k$'s verify the above $\limsup$'s, so we have done.