This is an exercise in Rudin's Functional Analysis.
Let $X$ be a Banach space and let $Y$ be a closed subspace of $X$. Let $x_0^*\in X^*$, $\mu=\sup\{|x_0^*(x)|:x\in Y, \|x\|\leq 1\}$ and let $\delta=\inf\{\|x^*-x^*_0\|:x^*\in Y^{\perp}\}$. I want to show that $\mu=\delta$. Also I need to show that there exists $x^*\in Y^{\perp}$ such that $\delta=\|x^*-x_0^*\|$.
I proceed in this way:
Let $x\in Y$ and $\|x\|\leq 1$. Then for all $x^*\in Y^{\perp}$, $|x_0^*(x)|=|x^*(x)-x^*_0(x)|\leq \|x^*-x_0^*\|\|x\|\leq \|x^*-x_0^*\|$. Thus $\mu\leq \delta$. But how to show that $\delta\leq \mu$? Any suggestion is appreciated.
If $y^{\ast} := x_0^{\ast}\lvert_Y$, then $y^{\ast}\in Y^{\ast}$ and $$ \mu = \|y^{\ast}\| $$ By Hahn-Banach, $\exists z^{\ast} \in X^{\ast}$ such that $z^{\ast}\lvert_Y = y^{\ast}$ and $\|z^{\ast}\| = \|y^{\ast}\| = \mu$. But $x^{\ast} := x_0^{\ast}-z^{\ast} \in Y^{\perp}$, so $$ \delta \leq \|x^{\ast} - x_0^{\ast}\| = \|z^{\ast}\| = \mu $$