It is well known that given an operator $T:\ell_\infty\to\ell_\infty$ such that $Tx=0$ for each $x\in c_0$ there exists an infinite subset $M$ of the positive integers so that $Tx=0$ for each $x\in \ell_\infty(M)$. See Theorem 2.5.4 in the book on Banach spaces by Albiac and Kalton.

I am interested in a version of this result for $T:\ell_\infty(\Gamma)\to \ell_\infty(\Gamma)$ when $\Gamma$ is uncountable.

From $Tx=0$ for each $x\in c_0(\Gamma)$ I would like to derive the existence of a subset $\Gamma'$ of $\Gamma$ so that $Tx=0$ for each $x\in \ell_\infty(\Gamma')$ with card($\Gamma'$)=card($\Gamma$), or at least with $\Gamma'$ uncountable.

($Q_1$) Is this result true?

To extend the argument in the proof of the countable case, I need the existence of a family $(\Gamma_i)_{i\in I}$ with card$(I)>$ card$(\Gamma)$ of almost disjoint subsets ($\Gamma_i\cap\Gamma_j$ finite for $i\neq j$) of $\Gamma$ with card$(\Gamma_i)=$ card$(\Gamma)$ for each $i$, or at least with all the $\Gamma_i$ uncountable.

($Q_2$) Is it possible to find such a family, at least with some restrictions on card($\Gamma$)?