One counterexample is so simple as to be trivial:
The free particle has a completely continuous energy spectrum, since the Hamiltonian $H = \frac{p^2}{2m}$ has $[0,\infty)$ as its spectrum (this follows directly from $p$ having completely continuous spectrum $(-\infty,\infty)$. The reason this does not violate the spectral theorem/the countability of the basis of the Hilbert space is that this $H$ is an unbounded operator, and the "eigenstates" $\lvert p \rangle$ (which are $\psi_p(x) = \mathrm{e}^{\mathrm{i}px}$ in the position wavefunction representation) are not inside the Hilbert space (just note that $\psi_p(x)$ is not square-integrable on $\mathbb{R}$ to see that it is not in the canonical space of position wavefunctions $L^2(\mathbb{R},\mathrm{d}x)$). Only wavepackets, i.e. square-integrable superpositions of the plane waves $\psi_p(x)$, lie inside the Hilbert space of states.
In fact, the "eigenstates" belonging to continuous eigenvalues are never "normalizable", cf. this phys.SE question, and thus never are vectors in the actual Hilbert space, but only in the larger space of the so-called rigged Hilbert space, cf. this phys.SE question
Another counterexample would be the hydrogen atom, where the energies above a certain threshhold are continuous, and are again essentially free states.
(1) Yes, the point spectrum is countable in your hypotheses: otherwise the operator would have an uncountable set of pairwise orthogonal vectors since eigenvectors of a self-adjoint operator with different eigenvalues are orthogonal.
This is impossible because, in every Hilbert space, every set of (normalized) orthogonal vectors can be completed into a Hilbert basis by Zorn's lemma and every Hilbert basis is countable if the space is separable.
(2) No, it is not necessarily true that the continuous spectrum is uncountable. You may have one only point in the continuous spectrum for instance. This is the case for the spectrum of the self-adjoint operator $1/H$, where $H$ is the Hamiltonian of the harmonic oscillator. The only point of the continuous spectrum is $0$.
COMMENT. However I do not believe that discrete is a really appropriate adjective for the point spectrum. My impression is that your idea of discrete involves the fact that the points are isolated. This is not the case for the point spectrum in general also if the Hilbert space is separable. You may have a point spectrum coinciding with rational numbers, which are dense in $\mathbb R$ as well known.
Indeed, there are other decompositions of the spectrum. Within a certain approach it is defined the so-called discrete spectrum as the part of point spectrum made of isolated eigenvalues whose eigenspaces are finite-dimensional.
If the Hilbert space is not separable, it is even possible to construct a self-adjoint operator whose point spectrum is the whole $\mathbb R$.
COMMENT 2. It is not necessary to introduce the notion of rigged Hilbert space to define notions of approximated eigenvalues and eigenvectors. Given a self-adjoint operator $A:D(A)\to H$ in the Hilbert space $H$, it is possible to prove that $\lambda \in \sigma_c(A)$ if and only if $\lambda$ is not an eigenvalue (in proper sense) and, for every $\epsilon>0$ there is $\psi_\epsilon \in D(A)$ with $||\psi_\epsilon||=1$ such that $||A\psi_\epsilon - \lambda \psi_\epsilon|| < \epsilon$.
Best Answer
Rigged Hilbert spaces have no special relationship to quantum field theory. We can talk about more general elements of "the" Hilbert space such as wave functions looking like distributions. They don't have a finite norm but they're still useful to talk about.
Truly physical states that may be realized in practice are normalizable - their norm may be chosen to be 1, in fact, so rigged Hilbert spaces represent a mathematically convenient but physically unacceptable extension of the Hilbert space.
Quite generally, we want the Hilbert space to be separable in QFT, too. After all, the Hilbert space of a QFT isn't much different from the direct sum of the $N$-particle Hilbert spaces in (non-relativistic) QM (summed over non-negative integers $N$) and those are separable.
In QFT, we may encounter superselection sectors, i.e. separable Fock-style (but interacting) Hilbert spaces built around a vacuum that isn't unique but depends on a parameter. In this case, the total Hilbert space is the direct sum over the (uncountably many) values of the parameters labeling the vacua ("moduli"). But it's still true that only one superselection sector is ultimately relevant for the description of real physical phenomena, so even in this case, one may say that the ultimate fully allowed Hilbert space is separable.
QFT gives us new ways to organize the states and define the operators on the Hilbert space but at the end, it's right to think about a QFT as another quantum mechanical theory with countable bases. The same remark applies to string theory.