The problem comes from Erwin Kreyszig's Introductory Functional Analysis with Applications, section 7.4, problem 4:
Let $T:l^2\mapsto l^2$ be defined by $y=Tx, x=(\xi_j), y=(\eta_j), \eta_j=\alpha_j \xi_j$, where $(\alpha_j)$ is dense in $[0,1]$. Find $\sigma_p(T)$ and $\sigma(T)$.
Here $l^2$ represents the normed space of all infinite sequence $(\xi_1,\xi_2,\ldots)$ satisfying $\sum_{i=1}^\infty\xi_i^2<\infty$; $\sigma_p(T)$ is the point spectrum of $T$, namely all $\lambda$ such that $\ker (T-\lambda I)\ne\{0\}$, and $\sigma(T)$ is the spectrum of $T$, i.e. those $\lambda $ such that $(T-\lambda I)^{-1}$ doesn't exist
When I am trying to solve this problem, I find something seeming inconsistent with the open mapping theorem:
For $\forall \lambda\in[0,1]-\{\alpha_j, j\in\mathbb{N}\}$, $T_\lambda=T-\lambda I$ is obvious a bounded linear operator on $l^2$, and it never maps nonzero $x\in l^2$ to zero since $\alpha_j-\lambda\neq0$ for every $j$. Thus the resolvent operator $R_\lambda(T)=(T-\lambda I)^{-1}$ exists. Since $l^2$ is a Banach Space, by the open mapping theorem(or the bounded inverse theorem), $R_\lambda(T)$ is a bounded linear operator.
But on the other hand, for these $\lambda$'s, since $\{\alpha_j\}$ is dense in $[0,1]$,$\forall N>0$, there exists some $\alpha_k$ such that $|\alpha_k-\lambda|<1/N$. So for $y=(\delta_{kj})$ (where $\delta_{kj} = \mathbb{I}\{j=k\}$, i.e. $y$ is the sequence whose $k$th item is $1$ and otherwise zero), $R_\lambda(T)y = (|\alpha_k-\lambda|^{-1}\delta_{kj})$. Therefore $R_\lambda(T)$ is unbounded since $N$ is arbitrary, which contradicts previous result.
I must have mistaken something when applying the open mapping theorem, but I cannot find it.
You assumed that $T_\lambda$ is surjective (when applying bounded inverse theorem). It is not true. In fact a small modification of your second argument shows precisely that $T_\lambda$ cannot be surjective.