You are correct and the proof looks sufficiently rigorous. However, $\|T\|_\text{op}$ can be computed exactly. That is, $\|T\|_\text{op}=\dfrac12$. To show this, let $z=(z_1,z_2,z_3,\ldots)\in\ell^\infty$. Then,
$$T(z)=\left(\frac{z_2}{2},\frac{z_3}{3},\frac{z_4}{4},\ldots\right)$$
so that
$$\big\|T(z)\big\|_{\infty}=\sup\left\{\frac{|z_k|}{k}\,\Big|\,k=2,3,4,\ldots\right\}\leq \sup\left\{\frac{\|z\|_\infty}{k}\,\Big|\,k=2,3,4,\ldots\right\}=\frac{\|z\|_\infty}{2}\,.$$
Note that the equality holds for $z=(0,1,0,0,0,\ldots)$. This implies $\|T\|_{\text{op}}= \dfrac{1}{2}$.
You can write an explicit solution $x\in\ell^\infty$ to $x=y+T(x)$. That is, $$x=(1-T)^{-1}y=\left(\sum_{k=1}^\infty\,\frac{y_k}{k!},\sum_{k=2}^\infty\,\frac{2!y_k}{k!},\sum_{k=3}^\infty\,\frac{3!y_k}{k!},\ldots\right)$$
Nonetheless, you did sufficient and good work. I was just making additional comments.
Your conclusion regarding the point spectrum is true, with the slight alteration that $\sigma_p (T)=\{\epsilon\}$ when $\epsilon > 1$, rather than $\geq 1$. To determine the residual and continuous spectra, we investigate the surjectivity of $T-\lambda I$ for $\lambda \neq \epsilon$.
Clearly $T$ is not surjective, so suppose $\lambda \neq 0$. We have $(T-\lambda I)(x) = ((\epsilon-\lambda)x_1,x_1-\lambda x_2,...) =y \implies x_1 = \frac{y_1}{\epsilon - \lambda}$, $x_{n+1} = \frac{x_{n}-y_{n+1}}{\lambda}$, so
$$x_{n} =\frac{x_1-\sum_{k=2}^{n} \lambda^{k-2} y_k }{\lambda^{n-1}}$$
In particular, if $y_k$ is nonzero for only finitely many $k \in \mathbb{N}$, so $\exists$ $N \in \mathbb{N}$ such that $y_n =0$ for $n>N$, then $\forall n>N$ $x_n=\frac{x_N}{\lambda^{n-N}}$. Choosing $y \in \ell^2$ such that $x_N \neq 0$ ( $y_k = \delta_{N,k}$ works), we have $x \in \ell^2 \iff |\lambda| > 1$. That is to say, $T-\lambda I$ is not surjective for $|\lambda| \leq 1$.
Further, this shows that for $|\lambda|>1$, im$(T-\lambda I)$ includes all sequences $y \in \ell^2$ with only finitely many nonzero terms. To show surjectivity of $T-\lambda I$ for $|\lambda|>1$, then, it suffices to show $x \in \ell^2$ when $y \in \ell^2$ and $y_1=0$. In this case, $x_n = -\sum_{k=2}^n \lambda^{k-n-1}y_k$, so
$|x_n|^2 \leq \sum_{k=1}^n |\lambda|^{2(k-n-1)}|y_k|^2$, and thus
$$\sum_{n=1}^N |x_n|^2 \leq \sum_{n=1}^N \sum_{k=1}^n |\lambda|^{2(k-n-1)}|y_k|^2 = \sum_{k=1}^N |y_k|^2 \sum_{n=k}^{N} |\lambda|^{2(k-n-1)} \\ = \sum_{k=1}^N |y_k|^2 \sum_{n=1}^{N+1-k} |\lambda|^{-2n} \leq \frac{1}{|\lambda|^2-1}\sum_{k=1}^N |y_k|^2 \leq \frac{\|y\|^2}{|\lambda|^2-1}$$
Showing $x \in \ell^2$, and hence that $\sigma(T)=\{|\lambda| \leq 1\} \cup \{\epsilon\}$. I'll leave it to you to determine the breakdown into the residual/continuous spectra, if that's of concern.
Best Answer
Hint: Notice that if $(y_n)_{n=1}^\infty$ is in the image of $T$, then $y_n=x_n+x_{n+1}$, for some sequence $(x_n)_{n=1}^\infty\in\ell ^2$, which implies $$\sum_{n=1}^k(-1)^{n+1}y_n=x_1+(-1)^{k+1}x_{k+1}$$In particular, for $k\rightarrow \infty$, the alternating sum $\sum_{n=1}^\infty(-1)^{n+1}y_n$ converges (to $x_1$ in this case). This is a necessary condition for being in the image of $T$.
Can you think of a sequence $(y_n)_{n=1}^\infty\in\ell^2$ such that $\sum_{n=1}^\infty(-1)^{n+1}y_n$ does not converge?
Edit: You can prove that $\sum^k_{n=1}(-1)^{n+1}y_n=x_1+(-1)^ny_{n+1}$ by induction. Starting with $k=1$ we have simply $$y_1=x_1+(-1)^{k+1}x_{k+1}=x_1+x_2$$which is true by assumption. Assuming the formula is proved for $k$, $$\sum^{k+1}_{n=1}(-1)^{n+1}y_n=\sum_{n=1}^k(-1)^{n+1}y_n+(-1)^{k+2}y_{k+2}=\\x_1+\underbrace{(-1)^{k+1}x_{k+1}+(-1)^{k+2}x_{k+1}}_{=0}+(-1)^{k+2}x_{k+2}=x_1+(-1)^{k+2}x_{k+2}$$