Trying to solve a problem, I have reduced it to the following, which I do not know if it is true or not, or how to prove it:
Let $Y$ be a closed subspace of $(l_{\infty}, ||\cdot||_{\infty})$, such that it is also a subset of $(l_{1}, ||\cdot||_{1})$. Let $(y_n)_{n\ge1}\in Y$ be a sequence with $y_n\rightarrow0$ as $n\rightarrow \infty$ wrt $||\cdot||_{\infty}$. Then $y_n\rightarrow0$ as $n\rightarrow \infty$ wrt $||\cdot||_{1}$.
The original problem is: Let $Y$ as above. Define $T:Y\rightarrow l_1$ by $T(x_1, x_2, \dots )= (x_2, x_3, \dots)$. I need to show that T is bounded. Using closed graph theorem I have reduced it to the above.
The inclusion $j \colon l_1 \to l_{\infty}$ is continuous: For $x \in l_1$ we have $$\lVert x\rVert_{\infty} = \max \: \{ \lvert x_n\rvert : n \in \mathbb{N}\} \leqslant \sum_{n \in \mathbb{N}} \lvert x_n\rvert = \lVert x\rVert_1 \tag{1}$$ (for $x \in l_1$, even $x \in c_0$, the supremum of $\lvert x_n\rvert$ is attained, hence we can use $\max$ here).
Therefore $j^{-1}(Y) = Y \cap l_1 = Y$ is closed in $l_1$. Thus $Y$ is a Banach space under the restriction of the $l_1$-norm as well as under the restriction of the $l_{\infty}$-norm to $Y$. These two norms are comparable [see $(1)$], hence by the open mapping theorem they are equivalent.
From the equivalence of the two norms, the desired conclusion follows immediately.
It is worth noting that such a subspace must be finite-dimensional (which yields another proof of the equivalence of norms). Indeed, analogous to the argument above, $Y$ is a closed subspace of $l_p$ for all $p < \infty$, in particular for $p = 2$. Now recall that $F \colon l_2(\mathbb{Z}) \to L^2(S^1,\sigma)$, where $\sigma$ is the normalised Haar measure, given by $$F(x) = t \mapsto \sum_{n \in \mathbb{Z}} x_n e^{int} \tag{2}$$ is an isometric isomorphism. It doesn't matter whether we identify $l_2(\mathbb{N})$ with $l_2(\mathbb{Z})$ via a bijection $\mathbb{N} \to \mathbb{Z}$ or view it as a subspace via the natural inclusion, either way $F(Y)$ is a closed subspace of $L^2(S^1,\sigma)$. And since $Y \subset l_1$, $(2)$ converges absolutely and uniformly for all $y \in Y$, whence $F(Y) \subset C(S^1) \subset L^{\infty}(S^1,\sigma)$.
By a theorem of Grothendieck (Theorem 5.2 in Rudin's Functional Analysis), $F(Y)$ is finite-dimensional. Since $F$ is injective, it follows that $Y$ is finite-dimensional.