Sequence of linear functionals on $\ell^{\infty}$ I am having trouble with this question.

Let $X$ be a normed space and let $T : X\to \ell^{\infty}$ be a bounded linear operator.
$a)$ Show that there is a sequence $(f_n)_n$ in $X^*$ so that $Tx=(f_n(x))$ for all $x \in X.$
$b)$ Suppose that $X$ is a subspace of a normed space $Y$. Show that there is a bounded linear operator $S: Y\to \ell^{\infty}$ so that $||S||=||T||$ and that $Sx = Tx$ for all $x \in X$

I am not sure how to start with question $a)$.
For $b)$, I tried to find a $P: \ell^{\infty}\to \Bbb{R}$ so that the composite function $PT: X\to \Bbb{R}$ is a bounded linear functional. Then I applied the Hahn-Banach extension theorem to say that there exists a $PS: Y\to \Bbb{R}$ which is a bounded linear functional such that $PS(x)=PT(x)$ for all $x \in X$. And then I got stuck.
Please advise me on what to do. Thank you.
 A: a. Let $f_n = \operatorname{proj}_n \circ T$ for all $n \in \Bbb{Z}$, where $\operatorname{proj}_n: \ell^\infty \to \Bbb{R}$ denotes the projection of a bidirectional sequence into its $n$-th component.  Clearly, $\operatorname{proj}_n$ is a bounded linear operator on $\ell^\infty$, so the composition $f_n = \operatorname{proj}_n \circ T$ of bounded linear operators is again a bounded linear operator.  This shows that $f_n \in X^*$.  By construction, $Tx = (f_n(x))_n$ for all $x \in X$.
b. Apply Hahn-Banach extension theorem on $f_n \in X^*$ for each $n \in \Bbb{Z}$ to obtain its extension $g_n \in Y^*$ so that $g_n|_{X} = f$ and $\lVert f_n \rVert = \lVert g_n \rVert$.
Construct $S:Y \to \ell^\infty$ componentwise.  i.e. $S(y) = (g_n(y))_{n \in \Bbb{Z}}$ for all $y \in Y$.  We need to show that $S$ is well-defined.  (The output of $S$ lies in $\ell^\infty$.)
$$\sup_{n \in \Bbb{Z}} |g_n(y)| \le \sup_{n \in \Bbb{Z}} \lVert g_n \rVert \lVert y \rVert = \sup_{n \in \Bbb{Z}} \lVert f_n \rVert \lVert y \rVert \le \sup_{n \in \Bbb{Z}} \lVert \operatorname{proj}_n \rVert \lVert T \rVert \lVert y \rVert = \lVert T \rVert \lVert y \rVert$$
Linearity of $S$ is evident from that of $g_n$, and the boundedness of $S$ follows from the above inequality.  $S|_X = T$ follows from $g|_X = f$. We have $\lVert S \rVert \le \lVert T \rVert$, and it remains to show the reverse inequality.  This is not hard from $S|_X = T$ and the definition of $\lVert \cdot \rVert_\infty$.
$$\lVert Tx \rVert = \lVert Sx \rVert \le \lVert S \rVert \lVert x \rVert \quad\forall\,x \in X$$
Since the domain of $T$ is the subspace $X$, we conclude that $\lVert T \rVert = \lVert S \rVert$. $\tag*{$\square$}$
