Getting the most general form of Mayer-Vietoris from the Eilenberg-Steenrod axioms I asked this question a while ago on MSE, got no answer, put a bounty on it, still got no answer, was advised to ask here instead, hesitated, forgot about the question for a while and now remembered it. I'm still not sure if it really qualifies as research level. I let you decide...

I'd like to derive the most general form of the Mayer-Vietoris sequence from the Eilenberg-Steenrod axioms for homology (in particular: I do not want to use the definition of $H_\ast(X)$ in terms of simplices). By that I mean the existence of an exact sequence of the form
$\cdots \to H_n(X_{12},A_{12})\to H_n(X_1,A_1)\oplus H_n(X_2,A_2)\to H_n(X,A) \to H_{n-1}(X_{12},A_{12})\to\cdots$
whenever $(A,A_1,A_2)\subseteq(X,X_1,X_2)$ are two excisive triads and $A_{12}:=A_1\cap A_2$, $X_{12}:=X_1\cap X_2$.
It is easy to do that in the special case $A=A_1=A_2$ by looking at the two long exact sequences for pairs from the inclusions $A\subseteq X_{12}\subseteq X_1$ and $A\subseteq X_2\subseteq X$ respectively. These two sequences form a Barratt-Whitehead ladder and the lemma of Mayer-Vietoris applies.
Basically the same approach works in the special case $X=X_1=X_2$.
Both of these proofs are well known in the literatur, but so far I was unable to find a proof for the general version either in the books or myself that did not go through the realisation of $H_\ast$ as the homology of some chain-complex generated by simplices.
All my previous attempts consisted of doodling one diagram after the other, but may be there is a simpler solution. After realising that $H_\ast(X,A) = \tilde{H}_\ast(C_A^X)$, where $C_A^X$ is the mapping cone of the inclusion $A\to X$, one could also ask whether $(C_A^X, C_{A_1}^{X_1}, C_{A_2}^{X_2})$ is an excisive triad. So I'm asking: Is it?
 A: If you are willing to work with mapping cones, then this follows from looking at the triple (= threefold iterated) mapping cone for the cube with vertices $A_{12} = A_1 \cap A_2$, $A_1$, $A_2$, $A$, $X_{12} = X_1 \cap X_2$, $X_1$, $X_2$ and $X$ in two different ways.
Let us use your notation $C_A^X = X \cup_A CA$, so that there is a natural isomorphism $H_*(X, A) \cong \tilde H_*(C_A^X)$.  If $(A, A_1, A_2)$ is excisive, then the double (= twofold iterated) mapping cone for the square with vertices $A_{12}$, $A_1$, $A_2$ and $A$ has the homology of a point.  Likewise, if $(X, X_1, X_2)$ is excisive, then the double mapping cone for the square with vertices $X_{12}$, $X_1$, $X_2$ and $X$ has the homology of a point.  Thus the triple mapping cone for the cube has the homology of a point.  This is homeomorphic to the double mapping cone for the square with vertices $C_{A_{12}}^{X_{12}}$, $C_{A_1}^{X_1}$, $C_{A_2}^{X_2}$ and $C_A^X$.  Since $C_{A_{12}}^{X_{12}} = C_{A_1}^{X_1} \cap C_{A_2}^{X_2}$, this shows that $(C_A^X, C_{A_1}^{X_1}, C_{A_2}^{X_2})$ is excisive, and gives you the exact Mayer-Vietoris sequence
$$
\dots \overset{\partial}\to H_n(X_{12}, A_{12}) \to H_n(X_1, A_1) \oplus H_n(X_2, A_2) \to H_n(X, A) \overset{\partial}\to \dots
$$
by the Barratt-Whitehead lemma.  (Note the spelling of Michael Barratt's name.)
