Problem:
Let $X$ and $Y$ be Banach spaces. Prove that the set $\Omega = \{T \in L(X,Y) \mid T \text{ is surjective}\}$ is an open set in the Banach space $L(X,Y)$ equipped with the operator norm.
What I Have Tried: I aim to show that if $T_0$ is a surjective operator such that for every $y \in Y$, there exists a solution $x_0$ satisfying $T_0x_0 = y$, then for any $T_1$ sufficiently close to $T_0$, there exists a solution $x_1$ satisfying $T_1x_1 = y$. I think the "Open mapping theorem" might be useful to find such a solution for $T_1x_1 = y$ from the original solution $x_0$, but i failed in constructing such "Approximate solution sequence". Any insights or guidance on this matter would be greatly appreciated.
Let $T \in L(X, Y)$ be surjective. Let $q: X \to X/\ker(T)$ be the natural quotient map. Observe that, by open mapping theorem, $S: X/\ker(T) \to Y$ given by $T = S \circ q$ is invertible.
Claim: Any $R \in L(X, Y)$ with $\|R - T\| < \frac{1}{2} \|S^{-1}\|^{-1}$ is surjective.
Proof: Let $y_0 \in Y$ be an arbitrary unit vector. It suffices to prove $y_0 \in \mathrm{range}(R)$. Since $T$ is surjective, we may choose $x_0 \in X$ with $Tx_0 = y_0$. Furthermore, by the definition of $S^{-1}$, we may choose $x_0$ s.t. $\|x_0\| \leq \frac{3}{2}\|S^{-1}\|\|y_0\|$. Proceed by induction. Assume $x_0, \cdots, x_n \in X$ and $y_0, \cdots, y_n \in Y$ have been chosen with $\|x_i\| \leq \frac{3}{2}\|S^{-1}\|\|y_i\|$ for all $0 \leq i \leq n$ and $\|y_{i+1}\| \leq \frac{3}{4} \|y_i\|$ for all $0 \leq i \leq n-1$. Let $y_{n+1} = (T-R)x_n$. Then $\|y_{n+1}\| \leq \|R - T\|\|x_n\| \leq \frac{3}{4}\|y_n\|$. Choose $x_{n+1} \in X$ s.t. $Tx_{n+1} = y_{n+1}$ and again, by definition of $S^{-1}$, we may assume $\|x_{n+1}\| \leq \frac{3}{2}\|S^{-1}\|\|y_{n+1}\|$.
Now, we observe that $\|x_n\| \leq \frac{3}{2}\|S^{-1}\|\|y_n\| \leq \frac{3}{2}\|S^{-1}\|(\frac{3}{4})^n \|y_0\| = \frac{3}{2}\|S^{-1}\|(\frac{3}{4})^n$. So $\sum_{n \geq 0} \|x_n\| < + \infty$. Thus, as $X$ is a Banach space, $x = \sum_{n \geq 0} x_n$ is well-defined. We have,
$$\begin{split} Rx &= \sum_{n \geq 0} Rx_n\\ &= \sum_{n \geq 0} Tx_n + \sum_{n \geq 0} (R-T)x_n\\ &= \sum_{n \geq 0} y_n + \sum_{n \geq 0} (-y_{n+1})\\ &=y_0 \end{split}$$
So $y_0 \in \mathrm{range}(R)$, as claimed. $\square$