If $U$ is unitary operator then spectrum $\sigma(U)$ is contained inside the unit circle 
In a Hilbert space, let $U$ be a continuous operator which it unitary. Prove $\sigma (U)\subseteq \Bbb{S}^1$.

It is important for me to know how I am doing, and I didn't come by a clear explanation so it would be appreciated to have your correcting and guiding.
Attempt: $U^{*}U=UU^{*}=I$ and therefore $U^{-1}=U$. I have
$$
\| U x \|^2
= \langle U x, U x \rangle
= \langle U^* U x, x \rangle
= \langle x, x \rangle
= \|x\|^2
$$
and the same goes for $U^*$ and therefore $\| U \| = \| U^{-1} \| = 1$. 
Looking at $U-\lambda$ I can take
$$
-\lambda U(U^{-1}-{1\over \lambda}I)
=U-\lambda I.
$$
Clearly, for $\lambda \ne 0$, $-\lambda U$ is defined and is a linear operator.
$(U^{-1}-{1\over \lambda}I)$ is invertible if $|\lambda|\ge 1$ and therefore $\sigma (A^{-1})=\sigma (A)\subseteq \{|z|\le1\}$. But from the RHS, $U-\lambda I$ is invertible if $|\lambda|\le 1$ and therefore $\sigma(A)\subseteq \Bbb{S}^1$. I have noticed my claims tend to have holes in them many times which I cannot spot. If it is substantially true, how can I make sure it is completely formal?
 A: In your proof slight modification is needed.
We have, $UU^* =U^*U =I$. So $U$ is invertible with $U^{-1}= U^*$. 
Also as you have already shown that $||U || = 1 = ||U^{-1}||$, we must have 
$$\sigma (U) \subset \{\lambda \in \mathbb{C} : |\lambda| \leq 1\}\;\; \text{and}\;\; 
\sigma (U^{-1}) \subset \{\lambda \in \mathbb{C} : |\lambda| \leq 1\}.$$
Now to show $\sigma (U) \subset \{\lambda \in \mathbb{C} : |\lambda| = 1\}, $ it is sufficient show that for $(U-\lambda I)$ is invertible for every $\lambda$ satisfying $|\lambda| <1$. As $U$ is already invertible, sufficient to show that  $(U-\lambda I)$ is invertible for every $\lambda$ satisfying $0 <|\lambda| <1$.
Now from the relation $-\lambda U (U^{-1} - \frac{1}{\lambda}I) = (U- \lambda I)$, It is clear that, for $\lambda \neq 0$, 
if  $(U^{-1} - \frac{1}{\lambda}I)$ is invertible then $(U - \lambda I)$ becomes invertible. 
Now note that If $0 <|\lambda| < 1,$ then $|\frac{1}{\lambda}| > 1$, then in that case $(U^{-1} - \frac{1}{\lambda}I)$ invertible and hence $(U- \lambda I)$ invertble. This gives us $\sigma (U) \subset \{\lambda \in \mathbb{C} : |\lambda| = 1\}. $
