compactness of a sequence space

Question

Sorry if this question might be not well-posed, I'm very very new to topology.
I have a compact set $S$ of sequences $(x_n)_{n\in\mathbb N}$ in $\mathbb R^n$ and those sequences are bounded, in the sense that $S\subset l_\infty$.
I have a family of continuous functions $(f_k)_{k\in\mathbb N}$ going from $S$ to $\mathbb R^d$.
For a given sequence $s\in S$ I can thus define the sequence $(y_k)_{k\in\mathbb N}$ in $\mathbb R^d$, where for any $k$, $y_k = f_k(s)$.
Moreover I know that $y_k=f_k(s)$ is bounded in the infinity norm for any sequence $s\in S$ and any $k\in\mathbb N$ and $\lim_{k\to\infty} \|f_k(s)\|_\infty \le \bar a$ for a given finite constant $\bar a \in \mathbb R$. How can I conclude (if possible) that the set
$$ R := \left\{(f_k(s))_{k\in\mathbb N} \in l_\infty \,\mid\, s\in S \right\} $$ is compact ?
I thought that maybe I can start showing that $R$ is bounded in the sense that any sequence $(f_k(s))_{k\in\mathbb N}\in R$ has a $l_\infty$ norm less or equal than a given constant and then show that any Cauchy sequence converges inside, however I'm quite confused about
thanks a lot