Inverse of Identity plus Compact Operator I have a linear bounded operator $T:L_2(0,1) \rightarrow L_2(0,1)$ of the form
$T=I+K,$ 
where $I$ is the identity operator and $K$ is a Hilbert-Schmidt integral operator and hence a compact operator.
I am wondering if the following assertion is true: Operator $T$ is invertible, that is, there exists an operator $T^{-1}:L_2(0,1) \rightarrow L_2(0,1)$ such that $T^{-1}T = TT^{-1}=I$.
My Approach: I use the following results to prove the above assertion.
The following two results are from the book An Invitation to Operator Theory-Abramovich:
Lemma 2.8: For a bounded operator $S:X \rightarrow Y$ between two Banach Spaces the following are equivalent:
i) $S$ is surjective and one-to-one,
ii) $S$ is bounded below and range of $S$ $R(S)=Y$.
Corollary 2.15:  For a bounded operator $S:X \rightarrow Y$ between two Banach Spaces the following are equivalent:
i) Range $R(S)$ is closed,
ii) There exists a constant $c>0$ such that for each $y \in R(S)$ there exists some $x \in X$ such that $y=Sx$ and $\|x\| \leq c \|y\|$.
Finally, I also use the result (Fredholm Alternative) from the book by Evans that the range of identity plus compact operator $R(I+K)$ is closed.
The following is my proof:
Proof: Since $K$ is compact, range of $T=I+K$, $R(T) \subseteq L_2(0,1)$ is closed. Therefore, we obtain from Corollary 2.15(ii) that the operator $T$ is surjective, i.e., for each $y \in R(T)$, there exists some $x \in L_2(0,1)$ such that $y=Tx$. Moreover, from Corollary 2.15(ii) we have that 
there exists a constant $c>0$ such that
$\|x\| \leq c\|Tx\|, \quad \forall x \in L_2(0,1)$. 
Therefore, $Tx=0 \Rightarrow x=0$, hence, $T$ is one-to-one.
Thus, we have shown that the operator $T$ is surjective and one-to-one. Therefore, we may apply Lemma 2.8 to conclude that $R(T)=L_2(0,1)$. Finally, using these facts, we may use the open mapping theorem we conclude that the inverse $T^{-1}$ exists. End of Proof
Please critique the proof. Moreover, if there is a reference where this result is stated, I would be very grateful. Thank you.

Your proposed statement is equivalent to saying that for any compact operator $K\in B(L^2(0,1))$, we have $-1\notin\sigma(K)$, which is clearly not true.  
To see this, pick an orthonormal basis $\{e_n\}$ for $L^2(0,1)$, let $\{\lambda_n\}$ be a sequence of complex numbers converging to $0$ with $\lambda_1=-1$, and define $K\in B(L^2(0,1))$ by
$$K\left(\sum_{n=1}^\infty\alpha_ne_n\right)=\sum_{n=1}^\infty\lambda_n\alpha_ne_n.$$
