Let $\lambda \notin \{0_{\mathbb C}\} \cup \sigma_p(T)$. Show that $\lambda$ is in the resolvent set $\rho(T)$ of $T \in \mathcal L(\ell^2(\mathbb C))$, where $$ T(x_n) = \left(\frac{x_n}{n}\right)\,. $$
The point spectrum $\sigma_p(T)$ was defined as \begin{equation} \sigma_p(T) = \{ \lambda \in \mathbb C \mid (\lambda I - T) \text{ is not injective} \} \end{equation} and the resolvent set $\rho(T)$ as \begin{equation} \rho(T) = \{ \lambda \in \mathbb C \mid (\lambda I - T)^{-1} \in \mathcal L(\ell^2(\mathbb C)) \}\,. \end{equation}
An attempt
By definition, for $\lambda$ to be in the resolvent set $\rho(T)$, the relation $$ \Vert(\lambda I - T)^{-1}x\Vert_2 \leq M\Vert x \Vert_2 $$ should hold for all $x \in \ell^2(\mathbb C)$ and some $M > 0$. By assumption, we know that for the $\lambda$ in question, $(\lambda I - T)$ is injective, so for any two distinct $x, y \in \ell^2(\mathbb C)$ we have $$ (\lambda I - T)x \neq (\lambda I - T) y \,. $$ By writing $x = (\lambda I - T) y$ for a suitable $y \in \ell^2(\mathbb C)$, the $2$-norm of our inverse image becomes $$ \Vert(\lambda I - T)^{-1} x\Vert_2 = \Vert(\lambda I - T)^{-1} (\lambda I - T) y\Vert_2 = \Vert y\Vert_2 \leq M \Vert y\Vert_2 $$ for $M = 1$.
Is this sufficient or am I missing something crucial, especially conceptually?
Let $\lambda\neq 0$ such that $(T-\lambda I)$ is injective and define $S:=(T-\lambda I)^{-1}$ Note that $S$ is given by $(Sy)_{n}=\frac{y_n}{\frac{1}{n}-\lambda}.$
Since $\lambda\not\in \{0\}\cup \{\frac{1}{n}\}_{n\in\mathbb{N}},$ we get that $$ d:=\inf_{n\in \mathbb{N}} |\frac{1}{n}-\lambda|>0 $$ and thus, $|(Sy)_n|\leq \frac{|y_n|}{d},$ which implies that $\| S(y)\|_{\ell^2}\leq \frac{\|y\|_{\ell^2}}{d} $. We conclude that $S$ is bounded and $\lambda \in \rho(T)$.