Let $\mathcal{B}$ be a real Banach space and $S$ a closed subspace of $\mathcal{B}$. I want to prove that for $f_0 \notin S$, there is a linear functional $l$ such that $l(f) = 0$ for all $f \in S$ and $l(f_0) = 1$. Furthermore, we may arrange so that $\|l\|_{\mathcal{B}^\ast} = 1/d(f_0,S)$.
To do this we consider the gauge function $p$ on $\mathcal{B}$ defined by $p(f) = d(f,S)/d(f_0,S)$. As $S$ is closed it is clear that $d(f_0,S) \neq 0$ so this is well-defined and furthermore $p(f) = 0$ for all $f\in S$. $p$ is semi-additive by the triangle inequality and I believe $d(\alpha f,S) = \alpha d(f,S)$ because $S$ is a subspace. Now consider the linear functional $l_0$ on $V_0 := \operatorname{Span} {f_0}$ defined by $l(\alpha f_0) = \alpha$. Then we have $l(v) \leq p(v)$ for all $v \in V_0$ and Hahn-Banach gives the existence of $l$ on $V$ so that $l(v) \leq p(v)$ for all $v \in \mathcal{B}$.
My question is: However, I am not able to arrange for $l$ to be so that $\|l\| = 1/d(f_0,S)$, how can we do this? I have also constructed such an $l$ using a different gauge function but it's not really relevant for it doesn't involve the distance function $d(f,S)$.
The construction yields $\lVert l\rVert = 1/d(f_0,S)$. On the one hand, since $d(f,S) \leqslant \lVert f\rVert$ for all $f\in\mathcal{B}$, we have
$$\lvert l(f)\rvert \leqslant p(f) = \frac{d(f,S)}{d(f_0,S)} \leqslant \frac{1}{d(f_0,S)}\lVert f\rVert \Rightarrow \lVert l\rVert \leqslant \frac{1}{d(f_0,S)}.$$
On the other hand, we have
$$\lVert l\rVert \geqslant \sup_{s\in S} \frac{\lvert l(f_0-s)\rvert}{\lVert f_0-s\rVert} = \sup_{s\in S} \frac{1}{\lVert f_0-s\rVert} = \frac{1}{\inf\limits_{s\in S} \lVert f_0-s\rVert} = \frac{1}{d(f_0,S)}.$$