By construction, every $\alpha_k$ is an eigenvalue of $T$, hence we have
$$K = \overline{\{\alpha_k : k\in\mathbb{N}\}} \subset \sigma(T).$$
On the other hand, if $\lambda\notin K$, then there is a $\delta > 0$ such that $\lvert\lambda - \alpha_k\rvert > \delta$ for all $k$, and then we find the inverse of $\lambda I - T$ explicitly:
$$\begin{align}
(\lambda I - T)\left(\sum_{n\in\mathbb{N}} y_n\cdot e_n\right) &= \sum_{n\in\mathbb{N}} x_n e_n\\
\iff \sum_{n\in\mathbb{N}} (\lambda - \alpha_n)y_n e_n &= \sum_{n\in\mathbb{N}} x_n e_n\\
\iff \bigl(\forall n\in \mathbb{N}\bigr)\bigl( (\lambda-\alpha_n)y_n &= x_n\bigr)\\
\iff \bigl(\forall n\in\mathbb{N}\bigr)\biggl(y_n &= \frac{1}{\lambda-\alpha_n}x_n\biggr),
\end{align}$$
so
$$(\lambda I - T)^{-1}\left(\sum_{n\in\mathbb{N}} x_n e_n\right) = \sum_{n\in\mathbb{N}} \frac{x_n}{\lambda-\alpha_n}e_n.$$
Since $\lvert\lambda - \alpha_n\rvert > \delta$ for all $n$, we have $\left\lvert\frac{1}{\lambda-\alpha_n}\right\rvert < \frac{1}{\delta}$, and so $(\lambda I - T)^{-1}$ is a globally defined continuous operator, hence $\lambda \notin \sigma(T)$, which yields the other inclusion $\sigma(T) \subset K$, altogether $\sigma(T) = K$.
You mentioned all the data you need.
You have,using that $T$ is bounded for the first equality, and the continuity of the inner product in the fourth equality,
\begin{align}
\|Tx\|^2&=\Big\|\sum_n\langle x,e_n\rangle\,Te_n\Big\|^2
=\Big\|\sum_n\langle x,e_n\rangle\,e_{n+1}\Big\|^2\\[0.3cm]
&=\sum_n|\langle x,e_n\rangle|^2=\|x\|^2.
\end{align}
So $T$ is an isometry. But it is not surjective, since $e_1$ is orthogonal to the range of $T$.
Best Answer
Isometric means $$ \|Tx\|=\|x\|, $$ for all $x\in H$. Equivalently, for all $x,y\in H$ $$ \|T(x+y)\|^2=\|(x+y)\|^2. $$ But $$ \|(x+y)\|^2=\langle x+y,x+y\rangle=\|x\|^2+2\langle x,y\rangle+\|y\|^2, $$ while $$ \|T(x+y)\|^2=\langle T(x+y),T(x+y)\rangle=\|Tx\|^2+2\langle Tx,Ty\rangle+\|Ty\|^2. $$ Thus, for all $x,y\in H$ $$ \langle Tx,Ty\rangle=\langle x,y\rangle. $$ Note that $\langle x,Ty\rangle=\langle T^*,y\rangle$, and the above becomes $$ \langle T^*Tx,y\rangle=\langle x,y\rangle\quad\text{or}\quad \langle (T^*T-I)x,y\rangle=0 $$ for all $x,y\in H$. In particular, for $y=(T^*T-I)x$, it becomes $$ 0=\langle (T^*T-I)x,(T^*T-I)x\rangle=\|(T^*T-I)x\|^2, $$ for all $x\in H$, and thus $T^*T=I$.