if three hermitian operators commute in pairs, are they all mutually diagonalizable? 
Let $A,B,C$ be hermitian operators, such that $[A,B]=[B,C]=[A,C]=0$.
  Does that mean we can find a set of mutual eigenstates?

I believe the answer is "Yes", since this was assumed during some derivation I'm reading through. If there's no degeneracy for one of the operators, it's quite straightforward. Otherwise, I'm not sure how to prove this, or whether it has anything to do with the hermiticity of the operators.
 A: If you are in a finite-dimensional complex vector space $V$, with any number of commuting self-adjoint operators $S,T,\ldots$, there is a basis consisting of simultaneous eigenvectors. The key point is that $T$ stabilizes the $S$-eigenspaces: for $v$ in the $\lambda$-eigenspace $V_\lambda$ of $S$, 
$$
S(Tv) \;=\; (ST)(v) \;=\; (TS)(v) \;=\; T(Sv) \;=\; T(\lambda\cdot v)
\;=\; \lambda\cdot Tv
$$
Thus, from the $S$-decomposition $V=\bigoplus_\lambda V_\lambda$, $T$ stabilizes each $V_\lambda$, and is still hermitian there, so decomposes each $V_\lambda$ into eigenspaces. Each of these "nested" eigenspaces consists of simultaneous eigenspaces for both $S,T$. A "downward induction" proves the analogous result for any finite number of operators. An extra trick, using the finite-dimensionality, gives the same result for an arbitrary set of commuting operators on a finite-dimensional space.
In the infinite-dimensional situation, if all the commuting self-adjoint operators are compact, then a similar discussion succeeds.
Even without compactness, it is still true (by the same computation) that commuting operators preserve each others' eigenspaces (if any).
