Let $L(A,B)$ bounded operators, $A$ and $B$ Banach . Then $S$ the set of onto operators is open. I have been trying to prove this fact, but I am having trouble. I cannot even understand it intuitively. What makes it impossible to make the image grow until it is the whole space?
 A: I claim that if $S$ is onto, and $\|T-S\|_{op}\leq \frac{1}{2\cdot \|S\|_{op}}$ then $T$ is onto (This means that there exists an open subset around $S$ of onto operators).
Before we prove this claim I need a lemma

Lemma (existance of almost orthogonal vectors): Let $X$ be a normed space and $M\subseteq X$ be a closed subspace, then for every $1>\varepsilon>0$ there is some $x_\varepsilon\in X$ with $\|x_\varepsilon\| = 1$ and $d(x_\varepsilon,M)>1-\varepsilon$ 

Proof of Lemma: Let $u\in X\backslash M$, then $d(u,M)=d>0$. By definition, there exists $v\in M$ with $\|u-v\|<\frac{d}{1-\varepsilon}$. Observe that $d(u-v,M) = d(u,M)$ (because $M$ is a subspace). Let $x_\varepsilon = \frac{u-v}{\|u-v\|}$ we have that $d(x_\varepsilon,M)$ = $\frac{1}{\|u-v\|}d(u-v,M) = \frac{d}{\|u-v\|}>1-\varepsilon$. This proves the lemma.
Proof of the main claim: Suppose by contradiction that $T$ is not onto, then $T(A)\subseteq B$ is a closed vector space. Therefore, by the lemma for $\varepsilon=1/2$ there exists an element $b\in B$ with $\|b\|_B=1$ and $d_B(b,T(A))>1/2$.
Now $S$ is onto, so there exists $a$ such that $S(a)=b$. Since $\|b\|_B=1$ we have that $\|a\|_A\geq 1/\|S\|_{op}$.
It follows that $\|S(a)-T(a)\|_B > 1/2$ and therefore $\|S-T\|_{op}\geq 1/2\cdot \|a\| \geq 1/(2\|S\|_{op})$
