This question had to parts.
$a)$ Calculate $‖f‖$.
$b)$ Determine all the Hahn-Banach extensions of $f$ to the space $c$.
My work:
$a)$ Let $x=(x_1,x_2,\ldots)$, then $||f(x)||=||(x_1)||\leq||x||_{\infty}$ . This means that $||f||\leq1$. Choose $x=(1,0,\ldots)$ then $||x||_{\infty}=1$ and $||f(x)||=1$. Which means that $||f||=1$
$b)$ is where I get stuck. I know that I can write $c=c_0\oplus \lambda(1,1\ldots)$. However I am unsure how to continue from here
You're off to a good start. By noting the direct sum decomposition of $ c $, you know that any extension is completely determined by $ f $ along with its value at $ \mathbf{w} = (1, 1, \ldots) $. Call this value $ f(\mathbf{w}) = k \in \mathbb{F} $. Then you can calculate $$ \begin{align} \bar{f}(\mathbf{x}) &= f(\pi_{c_0} \mathbf{x} + \pi_{c_0^\perp}\mathbf{x})\\ &= f(\pi_{c_0} \mathbf{x}) + \bar f(\pi_{c_0^\perp} \mathbf{x})\\ &= f(\mathbf{x} - (\lim{\mathbf{x}}) \mathbf{w}) + \bar f((\lim{\mathbf{x}}) \mathbf{w})\\ &=f(\mathbf{x} - (\lim{\mathbf{x}})\mathbf{w}) + \lim \mathbf{x}\ \cdot \bar f(\mathbf{w})\\ &=f(\mathbf{x} - (\lim{\mathbf{x}})\mathbf{w}) + k \lim \mathbf{x} \end{align} $$ Edit: I noticed from our calculations in the comments that there is only one value of $ k $ which yields an extension $ \bar f $ of norm no larger than that of $ f $. So this problem is more interesting than I at first realized.