How to show that every Killing vector field is a matter collineation? Various texts make this claim, but no proof is given.

Explicitly, let $L$ denote the Lie derivative. Suppose $L_X g_{ab} = 0$ for some vector field $X$, called a Killing vector field. Suppose that the stress-energy tensor $T_{ab}$ satisfies the Einstein field equations. Then, show that $L_X T_{ab} = 0$.

 A: Edit: Note: I have posted another proof of this in another question, here. Those who prefer coordinates may find it slightly more palatable.

I gather from your comments that you can do this if you have $\mathcal{L}_X\text{Ric} = 0$. Thus I will outline a somewhat more general result, assuming a certain identity connecting Killing vectors and Riemann curvature that's an exercise in many textbooks.
A Killing vector field $X$ has $R^a{}_{bcd}X^d = \nabla_c\nabla_bX^a$, i.e.,
$$\nabla_Z\nabla_YX - \nabla_{\nabla_ZY}X= R(Z,X)Y \equiv_{\text{def}} \nabla_Z\nabla_XY - \nabla_X\nabla_ZY - \nabla_{[Z,X]}Y\text{.}$$
Whenever torsion vanishes, substitution gives
$$\begin{eqnarray*}
\mathcal{L}_X(\nabla_ZY) &=& \nabla_X\nabla_ZY - \nabla_{\nabla_ZY}X\\
&=&\nabla_Z\nabla_XY - \nabla_Z\nabla_YX + \nabla_{[X,Z]}Y\\
&=&\nabla_Z(\mathcal{L}_XY) + \nabla_{\mathcal{L}_XZ}Y\text{.}\end{eqnarray*}  $$
You should be able to apply this formula to the covariant-derivative form of the Riemann tensor to show that
$$\mathcal{L}_X[R(Y,Z)W] = R(\mathcal{L}_XY,Z)W + R(Y,\mathcal{L}_XZ)W + R(Y,Z)\mathcal{L}_XW\text{,}$$
i.e., the Lie derivative of the Riemann tensor along a Killing vector vanishes. The corresponding result for the Ricci tensor follows trivially.
A: This is not a complete answer, but fills in some of the missing pieces Trimok asked about in the comments to Stan's answer:
Note that I did not verify that Stan's proof actually works.
Re 1)
$$
\begin{align}
\nabla_Z \nabla_Y X
&= Z^\lambda(Y^\mu X^\nu_{;\mu})_{;\lambda} \partial_\nu
\\&= Z^\lambda(Y^\mu_{;\lambda} X^\nu_{;\mu} + Y^\mu X^\nu_{;\mu;\lambda}) \partial_\nu
\\&= \nabla_{\nabla_ZY}X + Z^\lambda Y^\mu\nabla_{\partial_\lambda}\nabla_{\partial_\mu} X
\\&= \nabla_{\nabla_ZY}X + R(Z,X)Y
\end{align}
$$
where the last equality was assumed and is proven over there.
Now, there's another step that might not be obvious to all readers:
$$
\mathcal L_X(\nabla_YZ) = [X,\nabla_YZ] = \nabla_X\nabla_YZ - \nabla_{\nabla_YZ}X
$$
where the second equality is due to zero torsion, ie
$$
\nabla_AB - \nabla_BA - [A,B] = 0
$$
for arbitrary $A,B$ and in particular $A=X, B=\nabla_YZ$.
Re 2)
This step was left as an exercise for the reader - just compute the left-hand side of the last equation. The expression $\mathcal L_X(\nabla_YZ)$ holds for generic $Y,Z$ - that's just a bit of unfortunate naming.
Re 3)
That's the Leibniz rule for tensors with one term $\mathcal L_XR$ missing.
A: A slightly different and, perhaps, more simple proof follows. $\def\lie{\mathit{£}}$
For $K^a$ a Killing vector, we have (Kostant formula), 
$$\nabla_a \nabla_b K^c = -R_{bca}{}^d K^d.$$
Using this, we may prove that the covariant and the Lie derivatives commute:
$$\lie_K \nabla_a X^b = \nabla_a \lie_K X^b,$$ for arbitrary $X^a$. We have:
$$\lie_K(R_{abcd}X^d )= 2\nabla_{[a}\nabla_{b]} \lie_K X_c = R_{abcd}\lie_K X^d,$$ and $$\lie_K (R_{abcd}X^d) = \lie_K R_{abcd} X^d + R_{abcd} \lie_K X^d,$$ and these two combined yield $\lie_K R_{abcd} X^d = 0$ for arbitrary $X^a$, therefore $\lie_K R_{abcd} = 0$. Obviously, the Lie derivatives of Riemann tensor's contractions also vanish, therefore $\lie_K T_{ab} = 0$ by Einstein's equation.
