Bounded operator such that image of ball is $(1-\varepsilon)$-dense is surjective. This is not a homework problem, I found it in some notes and I am curious on how to prove it. The statement is as follows:
Let $T:X\to Y$ be a bounded linear operator between Banach spaces. If there is $\varepsilon>0$ and $R<\infty$ such that $T(B_X(R))$ is $(1-\varepsilon)$-dense in $B_Y(1)$ then $T$ is surjective.
Here $(1-\varepsilon)$-dense means that for every point in $B_Y(1)$ the ball of radius $(1-\varepsilon)$ intersects $T(B_X(R))$.
What I've thought so far:
It is enough to show that $T(B_X(R))$ contains some ball around $0\in Y$.
Assume that this is not the case so that for every $n$ the ball $B_Y(\frac{1}{n})$ is not contained in $T(B_X(R))$. By the $(1-\varepsilon)$-density we have that for any $y\in B_Y({\frac{1}{n}})\setminus T(B_X(R))$ there is some $z\in T(B_X(R))$ such that $d(y,z)<(1-\varepsilon)$. Furthermore, we have $d(0,z)\le d(0,y)+d(y,z)<\frac{1}{n}+(1-\varepsilon)$. If we let $n$ be large enough so that $(1-\varepsilon)+\frac{1}{n}<1$ then $z\in B_Y(1)$.
I don't know if this works or makes any sense and I don't know how to continue from here. Any help, hints or solutions would be appreciated.
 A: For simplicity adapt the statement to closed unit balls and $≤$ instead of $<$. Let $y\in Y$ lie in the unit ball and find some $z_0=T(x_0)\in T(B(R)_X)$ so that $\|y-z_0\|≤1-\epsilon$. Define $k_0:=\|y-z_0\|$. Then you've got a $z_1=T(x_1)\in T(B(R)_X)$ with
$$\|\frac{y-z_0}{k_0}-z_1\|≤1-\epsilon$$
which implies
$$\|y-z_0 - z_1\cdot k_0\|≤(1-\epsilon)k_0≤(1-\epsilon)^2$$
continue like this to get a sequences $z_n =T( x_n)$ and $k_n$ with:
$$\|y-z_0-k_0z_1-k_0k_1z_2-... -k_0\cdot...\cdot k_{n-1}z_n\|≤(1-\epsilon)^n$$
whence the sum
$$\sum_{n=0}^\infty z_n\prod_{i=0}^{n-1}k_i$$
converges to $y$. But each of the $k_i$ are less than $1-\epsilon$ and the $z_n$ are contained in $T(B(R)_X)$, so the norms of the summands are bounded by
$$\|T\|\,R\,(1-\epsilon)^{n-1}$$
and the sum converges absolutely. Indeed the sum
$$\sum_n x_n\prod_i k_i$$
converges in $X$ by the same argument. If you denote the limit with $x$ you then get with continuity of $T$:
$$T(x) = T(\sum x_n\prod k_i) = \sum T(x_n)\prod k_i = \sum z_n\prod k_i = y$$
and any $y$ in the unit ball of $Y$ admits a pre-image under $T$.
