Norm of a Toeplitz operator

Question

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I am going through the book called Introduction to large truncated Toeplitz matrices by Albrecht Böttcher & Bernd Silbermann and I am on the first chapter. I am trying to fill in the details of the proof of the following:

A bounded Toeplitz operator $T(a)$ with symbol $a\in L^\infty(\mathbb{T})$ has norm $\| T(a) \| = \| a \| _\infty$.

I understand why $\| T(a) \| \leq \| a \|_\infty$. The part I am stuck on is why is $\| T(a) \| \geq \| a \|_\infty$?

The book introduces a projection $S_n$, $n=1,2,3,\dots$: \begin{equation} (S_nx)_k = \begin{cases} 0, & k<-n, \\ x_k, & k\geq -n. \end{cases} \end{equation} The larger $n$ is, the fewer entries get turned into $0$ so $S_n$ converges strongly to the identity: for every square-summable $x$, \begin{equation} \|S_n x - Ix\| \xrightarrow{n\rightarrow\infty} 0. \end{equation}

So the next step is to take the bounded Laurent operator $L(a)$ and see that $S_n L(a) S_n$ converges strongly to $L(a)$, i.e., \begin{equation} \|S_n L(a) S_n x - L(a)x\| \xrightarrow{n\rightarrow\infty} 0. \end{equation} This is where I am stuck! How can I show this strong convergence?

I tried \begin{equation} \|S_n L(a) S_n x - L(a)x\| \leq \|S_n L(a) S_n x - L(a)S_nx\| + \|L(a)S_nx - L(a)x\| \end{equation} and if I let $y_n := L(a) S_n x$ the first term on the right-hand side of the inequality becomes \begin{equation} \|S_n L(a) S_n x - L(a)S_nx\| = \|S_n y_n - y_n \| \end{equation} but I don't think I can use the strong convergence as $y_n$ depends on $n$ ! So instead I tried \begin{equation} \|S_n y_n - y_n \| \leq \|S_n - 1 \| \|y_n\| \end{equation} but, $\|S_n - 1 \| = 1$.

Another thing I tried and was able to show is that \begin{equation} \|S_n L(a) S_m x - L(a)x\| \xrightarrow{m,n\rightarrow\infty} 0 \end{equation} but I'm not sure how to massage that any further into getting the result I want.

I am hoping someone here can provide some hints or advice. Thank you for your your help!