Prove that a subspace is dense in a Banach space 
Let $(X, \left\| \cdot \right\|)$ be a Banach space and let $E \subseteq X$ be a subspace with a constant $0<c<1$ such that
$$\inf_{a \in E}\left\| x - a \right\| \leq c \left\| x\right\|$$
for all $x \in X$. Prove that $E$ is dense in $X$.

So for each $x \in X$ and $\epsilon>0$, I need to show that the open ball $B_{\epsilon}(x)$ contains a point in $E$, but I don't see why the given inequality is enough to say this. We know that the closest point in $E$ to $x$ has distance at most the multiple of the norm of $x$ and I don't see a reason to have $c\left\|x \right\|< \epsilon$.
 A: Take $x\in X$ and $\varepsilon>0$. There is some $a_1\in E$ such that $\|x-a_a\|\leqslant c\|x\|$. And there is some $a_2\in E$ such that$$\|x-a_1-a_2\|\leqslant c\|x-a_1\|\leqslant c^2\|x\|.$$By the same idea, there is some $a_3\in E$ such that$$\|x-a_1-a_2-a_3\|\leqslant c^3\|x\|$$ and so on. Since $\lim_{n\to\infty}c^n\|x\|=0$, you can take $p\in\Bbb N$ such that $c^p\|x\|<\varepsilon$. And $\sum_{k=1}^pa_k\in E$.
A: A somewhat more abstract approach is as follows. Let $\overline{E}$ denote closure of $E$ in $X$.
Note that
$$\inf_{a\in E}\|x-a\|\geq\inf_{a\in\overline{E}}\|x-a\|$$
The r.h.s is the norm of the quotient space $X/\overline{E}$. By assumption, $\|x+\overline{E}\|_{X/\overline{E}}\leq c\|x\|$ for all $x\in X$, where $0<c<1$.
I claim that $X/\overline{E}$ is the trivial space, containing only zero. This of course proves that $\overline{E}=X$, so $E$ is dense in $X$.
Assume, on the contrary, that there exists a non-zero vector in the quotient space $X/\overline{E}$. Call it $x_0$.
The natural quotient map $Q:X\to X/\overline{E}$ defined by $Q(x)=x+\overline{E}$, maps the open unit ball of $X$
onto the open unit ball of $X/\overline{E}$.  Take some $t>0$ such that $y=tx_0\in X/\overline{E}$ satisfies $1>\|y\|>c$. There exists $x\in X$ such that $\|x\|<1$
and $Q(x)=y$. But then
$$c<\|y\|_{X/\overline{E}}=\|Q(x)\|_{X/\overline{E}}\leq c\|x\|<c,$$
a contradiction. Therefore there are no non-zero vectors in $X/\overline{E}$, hence $E$ is dense in $X$.
