Bounded Linear Operator + Surjective = Fried Egg 
Is there an easy way to show for bounded linear operators that happen to be surjective: $\exists r\geq0: \overline{B_1(0)}\subseteq T(B_r(0))$

Thanks for hints!
 A: 
so far I got: $B_\epsilon \subseteq \overline{T(B_1)}$ ...

The usual proof goes on to show that then - supposing Banach spaces - you have $B_\epsilon \subseteq T(B_2)$ (for example, one can obtain a smaller radius).
By homogeneity, you have
$$B_{c\cdot \epsilon} \subseteq \overline{T(B_c)}$$
for all $c > 0$. Now let $y \in B_\epsilon$ arbitrary. Since $T(B_1)$ is dense in $B_\epsilon$, there is an $x_0 \in B_1$ with $\lVert y - T(x_0)\rVert < \epsilon/2$. Let $y_1 = y - T(x_0)$. Since $T(B_{1/2})$ is dense in $B_{\epsilon/2}$, there is an $x_1 \in B_{1/2}$ with $\lVert y_1 - T(x_1)\rVert < \epsilon/4$. Let $y_2 = y_1 - T(x_1) = y - \bigl(T(x_0) + T(x_1)\bigr)$. Since $T(B_{1/4})$ is dense in $B_{\epsilon/4}$, there is ... You get the gist.
Since $X$ - let us suppose the domain of $T$ is named thus - is complete,
$$\xi = \sum_{n=0}^\infty x_n$$
exists, and $\lVert\xi\rVert \leqslant \sum\limits_{n=0}^\infty \lVert x_n\rVert < 2$. Show that $T(\xi) = y$.
Note that once you know that $T$ is open, you can deduce from $B_\epsilon \subseteq \overline{T(B_1)}$ that you actually have $B_\epsilon \subseteq T(B_1)$, since for every $\delta > 0$ the image $T(B_\delta)$ is a neighbourhood of $0$, whence
$$\overline{T(B_1)} \subseteq \bigcap_{\delta > 0}\left(T(B_1) + T(B_\delta)\right) = \bigcap_{\delta > 0} T(B_{1+\delta}),$$
so for every $\delta > 0$ you have $B_\epsilon \subseteq T(B_{1+\delta})$, which by homogeneity is equivalent to $B_{\epsilon/(1+\delta)}\subseteq T(B_1)$ for all $\delta > 0$, and hence
$$B_\epsilon = \bigcup_{\delta > 0}B_{\epsilon/(1+\delta)} \subseteq T(B_1).$$
A: I just remember a theorem which says:
If $g:E\to F$ is a homeomorphism such that $g^{-1}$ is lipschitz with $Lipg^{-1}<\lambda$ then $\overline{B_{\dfrac{r}{\lambda}}}(g(x))\subset g(\overline{B_r}(x))$. But i am not sure whether this can help you or not.
