Compactness in $l^\infty$

Yes, $A$ is compact. Note that $\ell^\infty$ is complete. Since $A$ is closed, it is complete.

Now, I shall prove that $A$ is totally bounded. Take $r>0$. Take $n\in\mathbb N$ such that $2^{-n}<c$. Let$$A_n=\left\{(a_1,\ldots,a_{n-1})\in\mathbb{R}^{n-1}\,\middle|\,(\forall i\in\{1,2,\ldots,n-1\}):|a_i|\leqslant 2^{-i}\right\}.$$Then $A_n$ is a compact subset of $\mathbb{R}^{n-1}$ (endowed with the $\sup$ metric). So, you can find a finite set $F\subset A_n$ such that the open balls centerered at the elements of $F$ with radius $r$ cover $A_n$. For each $(a_1,\ldots,a_{n-1})\in F$ consider the open ball centered at $(a_1,\ldots,a_{n-1},0,0,\ldots)$ with radius $r$. Then $A$ is contained in the union of these open balls.

Since $A$ is complete and totally bounded, it is compact.

Hint: Consider the endomorphism $T$ defined by $T(e_n)={{e_n}\over 2^n}$, and $T_n$ defined by $T(e_i)={{e_i}\over 2^i}$ for $i\leq n$, $T(e_i)=0, i>n$. Show that $T=lim_nT_n$, hence $T$ is compact. Show that $A$ is a closed subset of $T(B(0,2))$.

To continue the OP's approach: For the "diagonal subsequence" $(m_n)_{n\in\mathbb N}$ all coordinate sequences $a_k^{m_n}$ converge to $c_k$ and it is enough to show that then $a^{m_n} \to c=(c_1,c_2,c_3,\ldots)$ in $\ell^\infty$. Given $\varepsilon>0$ take $k_0\in \mathbb N$ with $2^{k_0-1}<\varepsilon$. For $k\ge k_0$ and all $n\in\mathbb N$ one has $|a_k^{m_n}-c_k|\le |a_k^{m_n}|+ |c_k| <\varepsilon$ and chosing $n_0$ such that $|a_k^{m_n}-c_k|\le \varepsilon$ for $n\ge n_0$ and $k\in\{1,\ldots,k_0-1\}$ one gets $\|a^{m_n}-c\|_\infty\le\varepsilon$ for $n\ge n_0$.