Let $Y \subsetneq X$ where $(X, \Vert \cdot\Vert)$ and $Y$ as a subspace is closed in $X$. Show that for $x_{0}\in X \setminus Y$ there exists $l \in X^{*}$ so that $l(x_{0})=\operatorname{dist}(x_{0},Y)$, $\Vert l\Vert=1$ and $l\vert_{Y}=0$.
First show that $\exists k \in (\operatorname{span}(Y\cup\{x_{0}\}))^{*}$ so that $k(x_{0})=\operatorname{dist}(x_{0},Y)$, $\Vert k\Vert=1$ and $k\vert_{Y}=0$ and then by Hahn Banach we extend this to the entire space.
My idea: define $k: \operatorname{span}(Y\cup\{x_{0}\})\ni t \mapsto\begin{cases} 0, t \in \operatorname{span}(Y)\\ \operatorname{dist}(t,Y), t \in \operatorname{span}(\{x_0\})\end{cases}$
Note that $k$ is linear on $\operatorname{span}(Y)$ since for any $\alpha \in \mathbb K$ and $y \in Y: k(\alpha y)=0=\alpha k(y)$
and linearity of $k\vert_{\operatorname{span}(\{x_{0}\})}$ from the linearity of $\operatorname{dist}(t,Y)$
now for boundedness, let $x \in \operatorname{span}(\{x_{0}\})$,
$\vert k(x)\vert=\operatorname{dist}(x,Y)\leq\Vert x-0\Vert=\Vert x\Vert\Rightarrow \Vert k\Vert\leq1$
But my real problem lies in showing that $\Vert k\Vert\geq1$. Any ideas on how to show this?
Also: Why is it necessary for $Y$ to be closed, I am not sure why it is relevant.
Note that $t\mapsto \mathrm{dist}(t, Y)$ for $t\in\mathrm{span}(x_0)$ is not linear (mapping $-x_0$ to the same nonzero number as $x_0$).
But, we can simply define $k(\alpha x_0+y)=\alpha\cdot\mathrm{dist}(x_0,Y)$ for $y\in Y$.
Observe that, with $z=\alpha x_0+y, \ \alpha\ne 0$, we have $$x_0-\frac z\alpha \in Y\ \implies\ \mathrm{dist}(x_0,Y)\le \|x_0-(x_0-\frac z\alpha)\| \\ \big|\frac{k(z)}\alpha\big|\le\big\|\frac z\alpha\|$$ Which implies $\|k\|\le1$.
The distance is an infimum, so there's a sequence $y_n\in Y$ such that $\|x_0-y_n\|\to\mathrm{dist}(x_0,Y)$.
Apply $k$ to $z_n=x_0-y_n$ to obtain $\|k\|\ge1$.