Existence of a normalized vector, far from proper closed subspace. Say we have a normed vector space $(E, \|\cdot\|)$, and a proper closed subspace $F\subsetneq E$, I want to prove the following:
$$(\forall \varepsilon > 0), (\exists e \in E) : \|e\|=1  \text{  and  }  d(e,F) \geq 1 - \varepsilon $$
Where $d(e,F)= \inf\limits_{y \in F} \|e-y\|$.
I started proving the following fact: 
Given $x \in E, \alpha \geq 1$, $\exists f \in F $ such that $\|x-f\|\leq\alpha \cdot d(x,F)$
Which is true by definition of $d$ as an infimum, because: $$ (\forall \varepsilon>0), (\exists f_{\varepsilon} \in F): d(x,F)\leq d(x,f_{\varepsilon})\leq d(x,F)+\varepsilon$$
And taking $\varepsilon = (\alpha - 1)\cdot d(x,F)$ we conclude.
So trying to use the last fact, let $\varepsilon >0,$ we define for $\bar{x} \notin F$ and $ \alpha \in \mathbb{R}_{+}$ to be determined, the $f\in F$ such that $ \|\bar{x}-f\|\leq \alpha\cdot d(\bar{x},F)$, then I don't know how to follow.
Mostly I don't know how to link $\bar{x}-f$ with the $e\in E $ that I'm trying to find, I could normalize, but I can't figure out where to get the inequality I'm looking for, my guess would be through fixing $\alpha = \frac{1}{1-\varepsilon}$ or something like that, but inequalities I manage to get are upper bounds.
Any help would be appreciated.

Since $d(x,F)>0$ for $x\notin F$ and $d(tx,F)=td(x,F)$ there exists $x_1$ with $d(x_1,F)=1$. Choose $a\in F$ with $||x_1-a||$ close to $1$. Now $x=x_1-a$ is almost the point you want...
