A step in the proof of the Spectral Theorem Let $T\in \mathcal{L}(H)$ be a compact self-adjoint operator on a separable Hilbert space $H$ over $\mathbb{R},$ and denote $||-||_{op}$ the operator norm; the context for the question is that I am reading a proof of the spectral theorem, where there are two things claimed without proof:

a) that given an eigenvalue $\lambda$ of $T,$ we have $\left| \lambda \right|\leq ||T||_{op};$
b) that at least one of $\text{inf}_{u\in H, ||u||=1}(Tu,u)$ or $\text{sup}_{u\in H, ||u||=1}(Tu,u)$ is an eigenvalue of $T.$

Can you show me a proof of these claims? Are these standard facts?
Edit
Since, as mentioned in the comments (which I thank), these are standard facts, can someone at least provide me a reference that contains a proof of point (b)? 
Edit '

I managed to prove that both $\text{inf}_{u\in H, ||u||=1}(Tu,u)$ and $\text{sup}_{u\in H, ||u||=1}(Tu,u)$ are actually contained in the spectrum $\sigma(T),$ but I know that not all elements of the spectrum are eigenvalues; so how does one prove that at least one of the two is an eigenvalue?

 A: If $T$ is a compact self-adjoint operator on a non-zero Hilbert space $H$ and
$$
m(T) = \sup\{\langle Tf,f\rangle:f\in H, \|f\|\leqslant 1\},
$$
then either $m(T)$ or $-m(T)$ is an eigenvalue of $T$. 
Note first that since $T=T^*$, we have for $f,g\in H$ $$\langle Tf, g\rangle = \langle f, Tg\rangle, $$
and setting $f=g$ we have $$\langle Tf, f\rangle = \overline{\langle f,Tf\rangle} = \langle f,TF\rangle,$$
so that $\langle Tf,f\rangle\in\mathbb R$, so that the absolute value $|\cdot|$ is not needed in the definition of $m(T)$. If $m(T) = 0$ then by the polarization identity we have for each $\|f\|=1$
$$
\langle Tf, f\rangle = \frac14(\|Tf+f\|^2 - \|Tf-f\|^2 + i\|Tf -if\|^2 -i\|Tf+if\|^2)= 0,
$$
from which
$$
4\mathrm{Re}(\langle Tf,f\rangle) = \|Tf+f\|^2 - \|Tf-f\|^2 = 0,
$$
hence
\begin{align}
\|Tf+f\|^2 = \|Tf-f\|^2 &\implies \langle Tf+f,Tf+f\rangle = \langle Tf-f,Tf-f\rangle\\
&\implies \|Tf\|^2 + \langle Tf,f\rangle + \langle f,Tf\rangle + \|f\|^2 = \|Tf\|^2  - \langle Tf, f\rangle - \langle f,Tf\rangle + \|f\|^2\\
&\implies 2\langle Tf,f\rangle = -2\langle Tf,f\rangle\\
&\implies \langle Tf,f\rangle = 0.
\end{align}
It follows that $T=0$.
Assume then that $|m(T)|>0$. Replacing $T$ by $-T$ as necessary, define $\varphi:H\to\mathbb R$ in $H^*$ by $\varphi(f) = \langle \varphi f,f\rangle$. Let $B = \{\psi\in H^*: \sup_{\|f\|=1}|\psi(f)| \leqslant 1\}$ be the closed unit ball in $H^*$. Since $H$ is separable, let $\{f_n\}$ be a countable dense subset. Then the following $\rho$ defines a metric on $B$:
$$
\rho(\varphi,\psi) = \sum_{n=1}^\infty 2^{-n}\frac{|\langle \varphi - \psi,\varphi_n\rangle|}{1+|\langle \varphi-\psi,\varphi_n\rangle|}.
$$
By the sequential Banach-Alaoglu theorem, $B$ is sequentially compact under $\rho$ - we may see this by a diagonalization argument similar to the one employed in the proof of the Arzelà–Ascoli theorem. Therefore by reflexivity of $H$, the closed unit ball $B$ is weakly compact. Also, the compactness of $T$ means that $T:(H,\text{ weak topology})\to (H,\text{ norm topology})$ is continuous, so $\varphi$ is continuous on $B$ equipped with the weak topology, and thus attains its maximum $m(T)$ at some $\psi\in B$. By maximality, $\|\psi\|=1$, which implies that $\psi$ maximizes the Rayleigh quotient $f\mapsto \frac{\langle Tf,f\rangle}{\langle f,f\rangle}$, from which we see that $T\psi = m(T)\psi$, thus proving the claim.
A: Because your operator is compact, its spectrum consists of eigenvalues, with the only possible exception of $0$. And if zero is not an eigenvalue, then the spectrum is a sequence that converges to zero. 
So if your sup and inf are distinct (and they always are unless $T=0$) then at most one of them can be zero, and so the other one is an eigenvalue. 
As for your first question, if $Tx=\lambda x$, then 
$$
\lambda=\lambda\,\frac{\langle x,x\rangle}{\|x\|^2}=\frac{\langle \lambda x,x\rangle}{\|x\|^2}=\frac{\langle Tx,x\rangle}{\|x\|^2}\leq\frac{\|Tx\|\,\|x\|}{\|x\|^2}\leq\frac{\|T\|\,\|x\|^2}{\|x\|^2}=\|T\|.
$$
