[proofplan]
We prove the equivalent resolvent assertion: every complex number off the unit circle gives an invertible operator $U-\lambda I$. The proof splits into the two regions $|\lambda|>1$ and $|\lambda|<1$, and in each region we factor $U-\lambda I$ into an invertible prefactor and an operator of the form $I-A$ with $\|A\|_{\mathcal L(H)}<1$. The inverse of $I-A$ is obtained directly from the Neumann series, and the two cases together show that no point outside the unit circle belongs to the spectrum.
[/proofplan]
[step:Record the Neumann series inverse used in both cases]
Let $A \in \mathcal L(H)$ satisfy $\|A\|_{\mathcal L(H)}<1$. Define $S \in \mathcal L(H)$ by the operator-norm convergent series
\begin{align*}
S := \sum_{n=0}^{\infty} A^n.
\end{align*}
The series converges because
\begin{align*}
\sum_{n=0}^{\infty} \|A^n\|_{\mathcal L(H)} \leq \sum_{n=0}^{\infty} \|A\|_{\mathcal L(H)}^n < \infty.
\end{align*}
For the partial sums $S_N := \sum_{n=0}^{N} A^n$, the telescoping identity gives
\begin{align*}
(I-A)S_N = I-A^{N+1}
\end{align*}
and
\begin{align*}
S_N(I-A)=I-A^{N+1}.
\end{align*}
Since $\|A^{N+1}\|_{\mathcal L(H)} \leq \|A\|_{\mathcal L(H)}^{N+1} \to 0$, passing to the operator-norm limit gives
\begin{align*}
(I-A)S = I
\end{align*}
and
\begin{align*}
S(I-A)=I.
\end{align*}
Thus $I-A$ is invertible in $\mathcal L(H)$ with inverse $S$.
[guided]
We will need the same elementary inverse construction twice, so we isolate it first. Let $A \in \mathcal L(H)$ be an operator with $\|A\|_{\mathcal L(H)}<1$. The candidate inverse for $I-A$ is the Neumann series
\begin{align*}
S := \sum_{n=0}^{\infty} A^n.
\end{align*}
This series is meaningful in $\mathcal L(H)$ because $\mathcal L(H)$ is a [Banach space](/page/Banach%20Space) under the operator norm and the norms of its terms are dominated by a convergent geometric series:
\begin{align*}
\sum_{n=0}^{\infty} \|A^n\|_{\mathcal L(H)} \leq \sum_{n=0}^{\infty} \|A\|_{\mathcal L(H)}^n < \infty.
\end{align*}
Hence the series converges in operator norm to some $S \in \mathcal L(H)$.
Now define the $N$th partial sum $S_N \in \mathcal L(H)$ by
\begin{align*}
S_N := \sum_{n=0}^{N} A^n.
\end{align*}
Multiplying out the finite sum gives the telescoping cancellation
\begin{align*}
(I-A)S_N = I-A^{N+1}.
\end{align*}
The same cancellation on the other side gives
\begin{align*}
S_N(I-A)=I-A^{N+1}.
\end{align*}
Because $\|A^{N+1}\|_{\mathcal L(H)} \leq \|A\|_{\mathcal L(H)}^{N+1}$ and $\|A\|_{\mathcal L(H)}<1$, we have $A^{N+1}\to 0$ in operator norm. Taking the operator-norm limit in both identities yields
\begin{align*}
(I-A)S = I
\end{align*}
and
\begin{align*}
S(I-A)=I.
\end{align*}
Therefore $S$ is both a left and right inverse for $I-A$, so $I-A$ is invertible in $\mathcal L(H)$.
[/guided]
[/step]
[step:Invert $U-\lambda I$ when $|\lambda|>1$]
Let $\lambda \in \mathbb C$ satisfy $|\lambda|>1$. Since $U$ is unitary, $U^*U=I$, and therefore
\begin{align*}
\|Ux\|_H^2 = (Ux,Ux)_H = (U^*Ux,x)_H = (x,x)_H = \|x\|_H^2
\end{align*}
for every $x \in H$. Hence $\|U\|_{\mathcal L(H)} \leq 1$. Define $A_\lambda \in \mathcal L(H)$ by
\begin{align*}
A_\lambda := \lambda^{-1}U.
\end{align*}
Then
\begin{align*}
\|A_\lambda\|_{\mathcal L(H)} \leq |\lambda|^{-1}\|U\|_{\mathcal L(H)} \leq |\lambda|^{-1}<1.
\end{align*}
By the Neumann series argument from the previous step, $I-A_\lambda$ is invertible in $\mathcal L(H)$. Since $-\lambda I$ is invertible with inverse $-\lambda^{-1}I$, and
\begin{align*}
U-\lambda I = -\lambda(I-\lambda^{-1}U),
\end{align*}
the product formula for inverses gives
\begin{align*}
(U-\lambda I)^{-1} = (I-\lambda^{-1}U)^{-1}(-\lambda^{-1}I).
\end{align*}
Thus $U-\lambda I$ is invertible.
[/step]
[step:Invert $U-\lambda I$ when $|\lambda|<1$]
Let $\lambda \in \mathbb C$ satisfy $|\lambda|<1$. Since $U$ is unitary, $UU^*=I$ and $U^*U=I$, so $U$ is invertible with inverse $U^*$. Also $U^*$ is unitary, and the same norm computation applied to $U^*$ gives $\|U^*\|_{\mathcal L(H)}\leq 1$.
Define $B_\lambda \in \mathcal L(H)$ by
\begin{align*}
B_\lambda := \lambda U^*.
\end{align*}
Then
\begin{align*}
\|B_\lambda\|_{\mathcal L(H)} \leq |\lambda|\|U^*\|_{\mathcal L(H)} \leq |\lambda|<1.
\end{align*}
By the Neumann series argument, $I-B_\lambda$ is invertible in $\mathcal L(H)$. Using $UU^*=I$, we factor
\begin{align*}
U-\lambda I = U-\lambda UU^* = U(I-\lambda U^*).
\end{align*}
Both factors on the right are invertible, so $U-\lambda I$ is invertible, with
\begin{align*}
(U-\lambda I)^{-1} = (I-\lambda U^*)^{-1}U^*.
\end{align*}
[/step]
[step:Conclude that the spectrum is contained in the unit circle]
Let $\lambda \in \mathbb C$ satisfy $|\lambda|\ne 1$. If $|\lambda|>1$, the second step shows that $U-\lambda I$ is invertible. If $|\lambda|<1$, the third step shows that $U-\lambda I$ is invertible. Therefore every $\lambda \in \mathbb C$ with $|\lambda|\ne 1$ lies outside $\sigma(U)$.
By the definition of the spectrum,
\begin{align*}
\sigma(U)=\{\lambda \in \mathbb C : U-\lambda I \text{ is not invertible in } \mathcal L(H)\}.
\end{align*}
Hence no point outside the unit circle belongs to $\sigma(U)$, which is exactly
\begin{align*}
\sigma(U) \subset \{z \in \mathbb C : |z|=1\}.
\end{align*}
This also proves the stated equivalent assertion that $U-\lambda I$ is invertible whenever $|\lambda|\ne 1$.
[/step]
