Theorem: Bounded Linear Operator on Space of Bounded Sequences Vanishing on Zero-Limit Sequences also Vanishes on Bounded Sequences with certain Support

Let $\GF \in \set {\R, \C}$.
Let $\struct {\map {\ell_\infty} {\N, \GF}, \norm {\, \cdot \,}_\infty}$ be the space of bounded sequences.
Let $\struct {\map {c_0} {\N, \GF}, \norm {\, \cdot \,}_\infty}$ be the space of zero-limit sequences. 
Let $T : \map {\ell_\infty} {\N, \GF} \to \map {\ell_\infty} {\N, \GF}$ be a bounded linear transformation with:
:$T \xi = {\mathbf 0}_{\ell_\infty}$ for each $\xi \in \map {c_0} {\N, \GF}$
Then there exists a countably infinite set $A \subseteq \N$ such that:
:$T \xi = {\mathbf 0}_{\ell_\infty}$ 
for each $\xi \in \map {\ell_\infty} {A, \GF}$, where $\map {\ell_\infty} {A, \GF}$ denotes the space of bounded sequences supported on $A$.

Proof:
Let $\struct {\map \BB {\map {\ell_\infty} {\N, \GF} }, \norm {\, \cdot \,}_{\map \BB {\ell_\infty} } }$ be the space of bounded linear transformations $\map {\ell_\infty} {\N, \GF} \to \map {\ell_\infty} {\N, \GF}$. 
From Countably Infinite Set has Uncountable Family of Subsets with Finite Intersection, there exists an uncountable set $I$ and an $I$-indexed family of sets $\family {S_i}_{i \mathop \in I}$ such that:
:$S_i \cap S_j$ is finite for each $i, j \in I$ with $i \ne j$.
{{AimForCont}} that for all $i \in I$, there exists $\xi_i \in \map {\ell_\infty} {S_i, \GF}$ such that $T \xi_i \ne {\mathbf 0}_{\ell_\infty}$.
By replacing $\xi_i$ with $\dfrac {\xi_i} {\norm {\xi_i}_\infty}$ if necessary, suppose that $\norm {\xi_i}_\infty = 1$ for all $\alpha \in I$.
For each $n \in \N$, set:
:$I_n = \set {\alpha \in I : \map {\paren {T \xi_i} } n \ne 0}$
For each $\alpha \in I$, there exists $n \in \N$ such that $\map {\paren {T \xi_i} } n \ne 0$, and so we have:
:$\ds \bigcup_{n \mathop = 1}^\infty I_n = I$
{{AimForCont}} that $I_n$ is countable for each $n \in \N$.
Then from Countable Union of Countable Sets is Countable, $I$ would be countable.
Hence there exists $N \in \N$ such that $I_N$ is uncountable.
Set:
:$I_{N, m} = \set {\alpha \in I_N : \cmod {\map {\paren {T \xi_i} } N} \ge m^{-1} }$
If $\map {\paren {T \xi_i} } N \ne 0$, then there exists $m \in \N$ such that $\cmod {\map {\paren {T \xi_i} } N} \ge m^{-1}$.
It follows that:
:$\ds \bigcup_{m \mathop = 1}^\infty I_{N, m} = I_N$
{{AimForCont}} that $I_{N, m}$ is uncountable for each $m \in \N$. 
Then from Countable Union of Countable Sets is Countable, $I_N$ would be countable.
This contradicts that $I_N$ is uncountable.
Hence there exists $M \in \N$ such that $I_{N, M}$ is uncountable.
For brevity, relabel $I = I_{N, M}$.
For each $\alpha \in I$, take $\alpha_i \ne 0$ such that:
:$\alpha_i \map {\paren {T \xi_i} } N = \cmod {\map {\paren {T \xi_i} } N}$
Now let $F$ be a finite subset of $I$ with $k$ elements.
Let:
:$\ds y = \sum_{i \mathop \in F} \alpha_i \xi_i$
For each $n \in \N$, let $J_n$ be the indices of those $i \in I$ such that $\map {\xi_i} n \ne 0$.
We then have:
:$\ds \map y n = \sum_{i \mathop \in J_n} \alpha_i$
Note that if $\size {J_n} > 1$, there exists $i, j \in I$ with $i \ne j$ and $\map {\xi_i} n \ne 0$ and $\map {\xi_j} n \ne 0$.
Since $\xi_i \in \map {\ell_\infty} {S_i, \GF}$ and $\xi_j \in \map {\ell_\infty} {S_j, \GF}$, it follows that $n \in S_i$ and $n \in S_j$. 
Hence if $\size {J_n} > 1$, then:
:$\ds n \in \bigcup_{i, \, j \in F, \, i \ne j} \paren {S_i \cap S_j}$
Since $F$ is finite, this is a finite union.
Further, by assumption $S_i \cap S_j$ is finite for all $i, j \in I$. 
Hence from Finite Union of Finite Sets is Finite:
:$\ds \bigcup_{i, \, j \in F, \, i \ne j} \paren {S_i \cap S_j}$ is finite.
Define:
:$\ds u = \begin{cases}\map y n & \size {J_n} = 1 \\ 0 & \text{otherwise}\end{cases}$
and:
:$\ds v = \begin{cases}\map y n & \size {J_n} > 1 \\ 0 & \text{otherwise}\end{cases}$
Note that $y = u + v$. 
Since the set of $n \in \N$ such that $\size {J_n} > 1$ is finite, we have that $\map v n = 0$ for large enough $n \in \N$ and hence $v \in \map {c_0} \GF$.
Hence $T v = {\mathbf 0}_{\ell_\infty}$.
So $T y = T u$. 
Further, for each $n \in \N$ such that $\map u n \ne 0$, we have $\map u n = \alpha_n$ with $\cmod {\alpha_n} = 1$. 
Hence we have:
:$\ds \norm u_\infty = \sup_{n \mathop \in \N} \cmod {\map u n} \le 1$
Hence from Fundamental Property of Norm on Bounded Linear Transformation, we have:
:$\norm {T y}_\infty = \norm {T u}_\infty \le \norm T_{\map \BB {\ell_\infty} }$
Further, we have:
{{begin-eqn}}
{{eqn	| l = T y
		| r = \map T {\sum_{i \mathop \in F} \alpha_i \xi_i}
}}
{{eqn	| r = \sum_{i \mathop \in F} \alpha_i T \xi_i
		| c = {{Defof|Linear Transformation}}
}}
{{end-eqn}}
and hence:
:$\ds \map {\paren {T y} } N = \sum_{i \mathop \in F} \alpha_i \map {\paren {T \xi_i} } N = \sum_{i \mathop \in F} \cmod {\map {\paren {T \xi_i} } N}$
where we recall $N$ is such that:
:$\set {\alpha \in I : \cmod {\map {\paren {T \xi_i} } N} \ge M^{-1} }$ is uncountable. 
We then have:
{{begin-eqn}}
{{eqn	| l = \sum_{i \mathop \in F} \cmod {\map {\paren {T \xi_i} } N}
		| r = \cmod {\map {\paren {T y} } N}
}}
{{eqn	| o = \le
		| r = \sup_{n \mathop \in \N} \cmod {\map {\paren {T y} } n}
}}
{{eqn	| r = \norm {T y}_\infty
}}
{{eqn	| o = \le
		| r = \norm T_{\map \BB {\ell_\infty} }
}}
{{end-eqn}}
For each $i \in F \subseteq I$, we have:
:$\cmod {\map {\paren {T \xi_i} } N} \ge M^{-1}$
Hence we obtain:
:$k M^{-1} \le \norm T_{\map \BB {\ell_\infty} }$ 
so that:
:$k \le M \norm T_{\map \BB {\ell_\infty} }$
Note that $k$ was arbitrary, hence taking:
:$k > M \norm T_{\map \BB {\ell_\infty} }$
gives a contradiction.
Hence our original assumption, that:
:for all $i \in I$, there exists $\xi_i \in \map {\ell_\infty} {S_i, \GF}$ such that $T \xi_i \ne {\mathbf 0}_{\ell_\infty}$.
was false.
Hence there exists $i \in I$ such that for all $\xi \in \map {\ell_\infty} {S_i, \GF}$, we have $T \xi = {\mathbf 0}_{\ell_\infty}$.
By construction, $S_i$ is countably infinite.
Hence setting $A = S_i$, we have the demand. 
{{qed}}