First note that the Hilbert space tensor product $H_A\otimes H_B$ can be isometrically identified with the space of Hilbert-Schmidt operators from $H_A$ to $H_B$ via
$$
x\otimes y\mapsto \langle \,\cdot\,,x\rangle y.
$$
Every Hilbert-Schmidt operator $T\colon H_A\to H_B$ is compact and therefore has a singular value decomposition
$$
T=\sum_{n} s_n\langle\,\cdot\,,e_n\rangle f_n
$$
with ONBs $(e_n)$ of $H_A$, $(f_n)$ of $H_B$ and a positive sequence $(s_n)\in\ell^2$. Thus every $v\in H_A\otimes H_B$ has a representation of the form
$$
v=\sum_{n} s_n e_n\otimes f_n
$$
with $(e_n),(f_n),(s_n)$ as above.
The proofs can be found in virtually any introductory text on functional analysis.
I'll just treat the minimum, the maximum works the same way. The minimum of $\{\langle Tx,x\rangle: x\in D(T),\|x\|=1\}$ does not necessarily exist (the set might even be unbounded from below), but the infimum does always equal the infimum of the spectrum (which is either $-\infty$ or actually the minimum).
This is not hard to prove once you know the spectral theorem: There exists a projection-valued measure $E$ on $\mathbb R$ such that
\begin{align*}
D(T)&=\left\{\xi\in H :\int\lambda^2\,d\langle E(\lambda)\xi,\xi\rangle<\infty\right\},\\
T\xi&=\int\lambda\,dE(\lambda)\xi,\;\xi\in D(T)
\end{align*}
and $\sigma(T)$ is the support of $E$, that is, the set of all $\lambda\in\mathbb R$ such that $E((\lambda-\epsilon,\lambda+\epsilon))\neq 0$ for all $\epsilon>0$.
Let $(\lambda_n)$ be a decreasing sequence in $\sigma(T)$ such that $\lambda_n\to-\infty$ and let $(\xi_n)$ be a sequence of unit vector such that $E((\lambda_n-1/n,\lambda_n+1/n))\xi_n=\xi_n$. In particular, $\xi_n\in D(T)$ and
$$
\langle T\xi_n,\xi_n\rangle=\int_{\lambda_n-1/n}^{\lambda_n+1/n}\lambda\,d\langle E(\lambda)\xi_n,\xi_n\rangle\leq (\lambda_n+1/n)\to\inf\sigma(T).
$$
Thus $\inf_{\|\xi\|=1}\langle T\xi,\xi\rangle\leq \inf\sigma(T)$.
On the other hand,
$$
\langle T\xi,\xi\rangle=\int_{\sigma(T)}\lambda\,d\langle E(\lambda)\xi,\xi\rangle\geq \inf\sigma(T)\int d\langle E(\lambda)\xi,\xi\rangle=\inf\sigma(T)\|\xi\|^2.
$$
Note that even if $T$ is bounded from below, the set $\{\langle T\xi,\xi\rangle:\xi\in D(T),\,\|\xi\|=1\}$ does not necessarily have a minimum. For example, the infimum of this set is clearly $0$ if $T$ is the (bounded self-adjoint) operator on $\ell^2$ acting by $T(a_n)=(a_n/n)$, yet there is no $(a_n)\in \ell^2$ such that $\langle T(a_n),(a_n)\rangle=0$.
Best Answer
If you're working strictly within the context of an inner product space or Hilbert space, the statement you have learned is false. Even if you could identify certain "states" that you would call "eigenstates," those objects are not vectors in the space. They are some unknown, unspecified object in some unknown and extended space.
For example, start with the position operator $M$ on an interval $[0,1]$. This operator is multiplication by $x$, meaning $(Mf)(x)=xf(x)$. This is a perfectly legitimate selfadjoint linear operator on the Hilbert space $L^2[0,1]$, but it has no eigenvectors, meaning that there are no non-trivial functions $f \in L^2[0,1]$ for which $Mf = \lambda f$, regardless of the choice of $\lambda$. The operator $M$ has "continuous spectrum" but no eigenvectors or eigenvalues. There just are no such "eigenstate" objects.
Physicists typically introduce the $\delta$ function into such a space (which is logically inconsistent) and claim that $(M \delta_{y})(x) = x\delta(x-y)=y\delta_{y}$. For many reasons, on many levels, that is Mathematical non-sense, and cannot be salvaged within $L^2[0,1]$ in a logically consistent way. The axioms of Quantum use Hilbert spaces, but $\delta$ functions cannot live in spaces such as $L^2[0,1]$ because two functions that are equal a.e. in $L^2[0,1]$ are identical, which means that pointwise values have no meaning for elements of $L^2[0,1]$. Dirac knew this, but he liked the intuition.
John von Neumann, who was a contemporary of P.A.M. Dirac, created consistent ways of dealing with the selfadjoint operators of Physics through the Spectral Theorem, and this was available to Dirac at the time; but Dirac chose an intuitive presentation over logical consistency. And it wasn't because rigorous Mathematics was unavailable at the time to handle everything; the correct Math was there.
All of the Hilbert spaces of Quantum must be separable, meaning that the space contains a countable dense subset. The reason for this has to do with constructibility, and being able to bootstrap to general answers through finite approximations. $L^2[0,1]$ is a separable space. One of the consequences of having a separable space, is that an orthonormal basis of such a space is always finite or countably infinite. You cannot have mutually orthogonal objects that are indexed by an interval of the real line, for example. That's impossible in separable spaces. And, if you have a selfadjoint operator $M$ on a Hilbert space, and you have $Mf=\lambda f$ and $Mg = \mu g$ for $\lambda\ne \mu$, then $(f,g)=0$ must hold. So, a selfadjoint operator on a Hilbert space that is allowed in Quantum Mechanics cannot have more than a countable number of actual eigenvalues. That's a consequence of correct axiomatic systems for Quantum. You can have continuous spectrum, but that's different than having a continuum of eigenvalues. If you enlarge the space to allow for such things, then you lose the ability to approximate in a finite way, which disconnects the theory from a setting where finite approximation makes sense.
Sorry to disappoint you. The theorem you have--as stated--is not true. Correct spectral theory gives you essentially the same thing, but not exactly that.