[proofplan]
Let $K:=\sigma(T)$, let $I\in\mathcal L(H)$ denote the identity operator on $H$, and let $C^*(T,I)\subset \mathcal L(H)$ denote the unital $C^*$-subalgebra generated by $T$ and $I$. Use the unital continuous functional calculus $\Phi:C(K)\to C^*(T,I)$. The inclusion $\sigma(f(T))\subset f(K)$ follows by explicitly inverting $f(T)-\lambda I$ when $\lambda$ is not in the range of $f$. For the reverse inclusion, if $f(z_0)=\lambda$, then $f-\lambda$ has a zero and is not invertible in $C(K)$; the functional calculus isomorphism and spectral permanence for unital $C^*$-subalgebras then force $f(T)-\lambda I$ to be non-invertible in $\mathcal L(H)$.
[/proofplan]
[step:Set up the continuous functional calculus on the spectrum]
Let
\begin{align*}
K:=\sigma(T)\subset\mathbb C.
\end{align*}
Let $I\in\mathcal L(H)$ denote the identity operator on $H$, and let $C^*(T,I)\subset\mathcal L(H)$ denote the unital $C^*$-subalgebra generated by $T$ and $I$. Since $T\in\mathcal L(H)$ is bounded and normal, the continuous functional calculus for bounded normal operators gives a unital $*$-homomorphism
\begin{align*}
\Phi:C(K)\to C^*(T,I)\subset \mathcal L(H)
\end{align*}
such that $\Phi(\operatorname{id}_K)=T$ and $\Phi(1_K)=I$, where $\operatorname{id}_K:K\to\mathbb C$ is the coordinate function $\operatorname{id}_K(z)=z$ and $1_K:K\to\mathbb C$ is the constant function $1_K(z)=1$. We write
\begin{align*}
f(T):=\Phi(f).
\end{align*}
We use the standard form of the continuous functional calculus saying that $\Phi$ is an isometric unital $*$-isomorphism from $C(K)$ onto $C^*(T,I)$, and we use spectral permanence for unital $C^*$-subalgebras: if an element of $C^*(T,I)$ is invertible in $\mathcal L(H)$, then its inverse belongs to $C^*(T,I)$.
[/step]
[step:Invert $f(T)-\lambda I$ when $\lambda$ is outside the range of $f$]
Let $\lambda\in\mathbb C\setminus f(K)$. Define
\begin{align*}
h_\lambda:K\to\mathbb C
\end{align*}
by
\begin{align*}
h_\lambda(z):=f(z)-\lambda.
\end{align*}
Since $\lambda\notin f(K)$, $h_\lambda(z)\ne 0$ for every $z\in K$. Therefore the function
\begin{align*}
g_\lambda:K\to\mathbb C
\end{align*}
defined by
\begin{align*}
g_\lambda(z):=\frac{1}{f(z)-\lambda}
\end{align*}
is continuous on $K$. By multiplicativity and unitality of $\Phi$,
\begin{align*}
\Phi(g_\lambda)\bigl(f(T)-\lambda I\bigr)=\Phi(g_\lambda)\Phi(h_\lambda)=\Phi(g_\lambda h_\lambda)=\Phi(1_K)=I.
\end{align*}
Also,
\begin{align*}
\bigl(f(T)-\lambda I\bigr)\Phi(g_\lambda)=\Phi(h_\lambda)\Phi(g_\lambda)=\Phi(h_\lambda g_\lambda)=\Phi(1_K)=I.
\end{align*}
Thus $f(T)-\lambda I$ is invertible in $\mathcal L(H)$, so $\lambda\notin\sigma(f(T))$. Hence
\begin{align*}
\sigma(f(T))\subset f(K).
\end{align*}
[guided]
We prove the first inclusion by constructing the resolvent explicitly. Fix $\lambda\in\mathbb C\setminus f(K)$. This means that no point of the compact set $K=\sigma(T)$ is sent to $\lambda$ by $f$, so the [continuous function](/page/Continuous%20Function)
\begin{align*}
h_\lambda:K\to\mathbb C
\end{align*}
given by
\begin{align*}
h_\lambda(z):=f(z)-\lambda
\end{align*}
has no zeros on $K$.
Because $h_\lambda$ is continuous and never zero, its reciprocal is again continuous. Define
\begin{align*}
g_\lambda:K\to\mathbb C
\end{align*}
by
\begin{align*}
g_\lambda(z):=\frac{1}{f(z)-\lambda}.
\end{align*}
Then $g_\lambda h_\lambda=1_K$ and $h_\lambda g_\lambda=1_K$ as functions in the commutative algebra $C(K)$.
Now apply the continuous functional calculus. The point of using the functional calculus is that it preserves products and constants. Since $\Phi$ is multiplicative and unital,
\begin{align*}
\Phi(g_\lambda)\bigl(f(T)-\lambda I\bigr)=\Phi(g_\lambda)\Phi(h_\lambda)=\Phi(g_\lambda h_\lambda)=\Phi(1_K)=I.
\end{align*}
The same computation in the opposite order gives
\begin{align*}
\bigl(f(T)-\lambda I\bigr)\Phi(g_\lambda)=\Phi(h_\lambda)\Phi(g_\lambda)=\Phi(h_\lambda g_\lambda)=\Phi(1_K)=I.
\end{align*}
Thus $\Phi(g_\lambda)$ is a two-sided inverse for $f(T)-\lambda I$ in $\mathcal L(H)$. By the definition of the spectrum of a bounded operator, this proves $\lambda\notin\sigma(f(T))$. Since every $\lambda\notin f(K)$ is outside $\sigma(f(T))$, we have
\begin{align*}
\sigma(f(T))\subset f(K).
\end{align*}
[/guided]
[/step]
[step:Use a zero of $f-\lambda$ to force non-invertibility]
Let $\lambda\in f(K)$. Choose $z_0\in K$ such that $f(z_0)=\lambda$. With $h_\lambda:K\to\mathbb C$ defined by
\begin{align*}
h_\lambda(z):=f(z)-\lambda,
\end{align*}
we have $h_\lambda(z_0)=0$. Hence $h_\lambda$ is not invertible in $C(K)$, because any inverse $u\in C(K)$ would satisfy
\begin{align*}
1=(u h_\lambda)(z_0)=u(z_0)h_\lambda(z_0)=0.
\end{align*}
Suppose, for contradiction, that $f(T)-\lambda I=\Phi(h_\lambda)$ is invertible in $\mathcal L(H)$. Since $\Phi(h_\lambda)\in C^*(T,I)$, spectral permanence for the unital $C^*$-subalgebra $C^*(T,I)\subset\mathcal L(H)$ implies that $\Phi(h_\lambda)^{-1}\in C^*(T,I)$. Because $\Phi:C(K)\to C^*(T,I)$ is onto, there exists $u\in C(K)$ such that
\begin{align*}
\Phi(u)=\Phi(h_\lambda)^{-1}.
\end{align*}
Therefore
\begin{align*}
\Phi(u h_\lambda)=\Phi(u)\Phi(h_\lambda)=I=\Phi(1_K).
\end{align*}
Since $\Phi$ is injective, $u h_\lambda=1_K$, contradicting the non-invertibility of $h_\lambda$ in $C(K)$. Thus $f(T)-\lambda I$ is not invertible in $\mathcal L(H)$, so $\lambda\in\sigma(f(T))$. Therefore
\begin{align*}
f(K)\subset\sigma(f(T)).
\end{align*}
[/step]
[step:Combine the two inclusions]
The previous steps prove
\begin{align*}
\sigma(f(T))\subset f(K)
\end{align*}
and
\begin{align*}
f(K)\subset\sigma(f(T)).
\end{align*}
Since $K=\sigma(T)$, these two inclusions give
\begin{align*}
\sigma(f(T))=f(\sigma(T)).
\end{align*}
This is the desired spectral mapping identity for the continuous functional calculus.
[/step]
