I have the linear operator $T: \ell^{\infty} \rightarrow \ell^{\infty}$ as $T \left( x_1, x_2, x_3, \cdots \right) = \left( x_2, x_3, \cdots \right)$. Then, $T$ is bounded and we have $\| T \| \leq 1$. Therefore, I can say that the spectrum is contained in the disc of radius $1$.
So, I take $\left| \lambda \right| > 1$ and I try to prove that this is a regular value. First, I want to prove that $T_{\lambda} = T - \lambda I$ is invertible (from its image). So, it is enough to prove that $T_{\lambda}$ is invertible. We see that $T_{\lambda} \left( x_i \right) = \left( x_{i + 1} - \lambda x_i \right)$. Now, if $T_{\lambda} \left( x_i \right) = 0$, then for all $i \in \mathbb{N}$, we have $x_{i + 1} = \lambda x_{i} = \lambda^i x_1$. So, if $x_1 \neq 0$, then the sequence $\left( x_i \right)$ is unbounded since $\left| \lambda \right| > 1$. Thus, we must have $x_1 = 0$ and consequently, $\left( x_i \right) = 0$.
That is, for $\left| \lambda \right| > 1$, $T_{\lambda}$ is invertible. Now, I wish to show that the image of $T_{\lambda}$ is dense in $\ell^2$ and the inverse is bounded. However, I cannot see how to prove the density of the image? I have a hunch that the image will be whole of $\ell^2$, but I cannot see a way to prove it.
As mentioned by Arctic Char in the comment, we already know that the spectrum is contained in the unit disc. Now, let us consider $\left| \lambda \right| \leq 1$. Then, the sequence $\left( 1, \lambda, \lambda^2, \cdots \right) \in \ell^{\infty}$ and $T_{\lambda} \left( 1, \lambda, \lambda^2, \cdots \right) = 0$. Therefore, $T^{\lambda}$ is not invertible for $\left| \lambda \right| \leq 1$ and we have that the spectrum is whole of the unit disc.
In fact, all the elements of the unit disc are eigenvalues of $T$!