Must every isometry have an associated Killing vector? I understand that the flows of Killing vector fields are isometries, and that one-parameter groups of isometries have an associated Killing vector which generates them, but are your Killing vectors guaranteed to give you all possible isometries of the manifold? I guess what I'm trying to ask is, do isometries necessarily come in families, or can you have "isolated" isometries which can't have an associated Killing vector field?
 A: Yes, you can have such "isolated" isometries.
Consider the real line $\mathbb R$ and the inversion mapping $x\to -x$.  This isometry does not arise from a killing vector because it's not "continuously connected to the identity."
A: The group of isometries of a given connected smooth (semi) Riemannian manifold is always a Lie group. However, a Lie group can include subgroups of discrete isometries that, barring the identity,  cannot be represented by continuous isometries and thus they have no Killing vectors associated with them. (Actually, only some elements of the connected component including the identity can be associated to Killing fields.)
For instance, referring to $\mathbb R^3$ equipped with the standard metric, the Lie group of isometries is the semidirect product of space translations  $\mathbb R^3$ and rotations $O(3)$ around a fixed point. The second mentioned subgroup of isometries, $O(3)$, admits a discrete subgroup: $\{I, -I\}$. The spatial inversion $-I$ cannot be associated with any Killing field. Similarly all the symmetries in $-I(SO(3))$ cannot be associated with Killing fields. 
