Subspace of a Normed Linear Space

1. Let $x \in X \setminus V$, and let us consider $W = V \oplus \langle u_0 \rangle$. Since $V$ is closed, $W$ is closed as well. Now, for $w = v + au_0 \in W$, we define $f(w) = a d_X(u_0, V)$.

Now, $f$ is a linear functional on $W$, $f(u_0) = f(0 + 1u_0) = d_X(u_0,V)$, and obviously for $v \in V$ we have $f(v) = f(v + 0u_0) = 0 d_X(u_0,V) = 0$. We want to show that $\lVert f \rVert = 1$.

For $a \neq 0$, we have \[ \lvert f(au_0+v) \rvert = \lvert a \rvert d_X(u_0,V) \leq \lvert a \rvert \cdot \lVert u_0 - (-a)^{-1}v \rVert = \lVert au_0 + v \rVert \] where the inequality comes from the fact that \[ d_X(u_0,V) = \min_{u \in V} \lVert u_0 - u \rVert \leq \lVert u_0 - (-a)^{-1}v \rVert. \]

By Hahn-Banach, $f$ extends to a linear functional $T \in X^\ast$ such that $\lVert T \rVert = 1$, $T(v) = 0$ for $v \in V$, and $T(a u_0) = a d_X(u_0, V)$, as desired.

2. Let $T$ such that $\lVert T \rVert = 1$ and $T = 0$ on $V$. For any $v \in V$, we have \[ T(u_0) = T(u_0) + T(v) = T(u_0 + v) \leq \lVert T \rVert \lVert u_0 + v \rVert = \lVert u_0 + v \rVert \] so \[ T(u_0) \leq \min_{v \in V} \lVert u_0 + v \rVert = d_X(u_0, V) \] as desired.