The space $X$, defined above, as well as the space $JT$ could not serve as counterexamples to the question I asked.More precisely the subspaces $X_*$, $JT_*$ are quasi-complemented in the corresponding space.We recall that a closed subspace $Y$ of $X$ is quasi-complemented if there exists a closed subspace $W$ of $X$ such that $Y \cap W = 0$ and $Y+W$ dense in $X$.Since the subspaces $X_*$, $JT_*$ are separable their quasi-complement is non separable thus for each one of them there is a non separable closed subspace with trivial intersection.This is a consequence of an old theorem due to H.P. Rosenthal https://www.sciencedirect.com/science/article/pii/0022123669900111 (Th.2.1)that states the following.
$Theorem$:If $Y$ is a closed subspace of $X$ such that $Y^\bot$ contains a reflexive subspace then $Y$ is quasi-complemented in $X$.
I think that the result remains valid if $Y^\bot$ contains $l_1$.
It holds that $JT_*^\bot$ is isometric to $l^{2} (2^N)$ and $X_*^\bot$ is isomorphic to $l^{1} (2^N)$ and this yields the quasi-complementation of both subspaces.
Hence my question remains open.The only known example of a subspace which is not quasi-complement is $c_0 (\Gamma)$ as a subspace of $l^\infty (\Gamma)$ with $\Gamma$ an uncountable set. This is a result of J. Lindenstrauss. However it is easy to see that there exists a non separable closed subspace $Y$ of $l^\infty (\Gamma)$ with $c_0 (\Gamma) \cap Y = 0$.