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?