In the section Axioms for homology from Hatcher's Algebraic Topology (page 161) he says:
Note that $\tilde{h}_n(x_0) = 0$ for all $n$, as can be seen by looking at the long exact sequence of reduced homology groups of the pair $(x_0,x_0)$.
I guess that he means the exact sequence of the second axiom, which gives the boundary map $\partial :\tilde{h}_n(X/A)\to\tilde{h}_{n-1}(A)$.
From this it is clear that $\tilde{h}_n(x_0)=\tilde{h}_{n-1}(x_0)$ for all $n$. But Hatcher is not assuming the dimension axiom, so I cannot say that $\tilde{h}_n(x_0)=0$ for $n\neq 0$, so how could I conclude that $\tilde{h}_n(x_0) = 0$ for all $n$?
I've thought that this could be shown using the first axiom if I could show that constant maps induce trivial maps in homology, but I don't see how to deduce this fact from the axioms because the proof I know for singular, cellular or simplicial homology uses that the homology of a point is trivial.
We have an exact sequence $$\tilde{h}_n(A)\stackrel{i_*}\to\tilde{h}_n(X)\stackrel{q_*}\to\tilde{h}_n(X/A)$$ where the first map is induced by the inclusion $i:A\to X$ and the second map is induced by the quotient map $q:X\to X/A$. But if $X$ and $A$ are both just a point, then $i$ and $q$ are both homeomorphisms, so $i_*$ and $q_*$ are both isomorphisms. But the image of $i_*$ is the kernel of $q_*$, so this implies $q_*=0$, and so $\tilde{h}_n(X)=0$ since $q_*$ is an isomorphism.