Let $X\ne \{0\}$ a Banach space, $\rho(T)$ is the resolvent of an operator $T$.
Proposition. Let $T\colon D(T)\to X$ a linear closed operator. We have that $$\lambda\in \rho(T)\iff(T-\lambda I)\quad\text{is bijective}$$
Proof. The implication $(\Leftarrow)$ is clear.
$(\Rightarrow)$ Since $(T-\lambda I)$ is continuous, exists $K\ge 0$ such that $\lVert(T-\lambda I)^{-1} \rVert\le K\lVert x\rVert$ for each $x=(T-\lambda I)y\in \text{Ran}(T-\lambda I).$
Then for all $y\in D(T)$, since $(T-\lambda I)y=x\in \text{Ran}(T-\lambda I), $ we have $$\lVert y\rVert=\lVert (T-\lambda I)^{-1}[(T-\lambda I)y]\rVert\le K\lVert (T-\lambda I)y\rVert\tag 1.$$
Since $\overline{\text{Ran}(T-\lambda I)}=X$, if $x\in X$, exists a sequence $\{y_n\}\subseteq D(T)$ such that $(T-\lambda I)y_n\to x.$
Question.$\{y_n\}$ is a Cauchy sequence.
My answer. We use the $(1)$ and the fact that the sequence $\{(T-\lambda I)y_n\}$ is a Caichy sequence, since it is convergent.
It's correct?