I've came across the following argument in Hatcher's "Algebraic Topology", as part of a proof of some version of the Künneth Formula.
In the following discussion, $h^*, k^*$ are both relative cohomology theories. $\mu$ is a natural transformation between those. We aim to show it's a natural isomorphism.
$X^n,X^{n-1}$ are the $n,{n-1}$ skeleta of a CW complex.
I'm quoting the argument that's bugging me - the full text can be found at Hatcher's web page - see Proposition $3.17$.
Let $\Phi : \coprod_\alpha (D_\alpha^n,\partial D_\alpha^n)\to (X^n,X^{n-1})$ be be a collection of characteristic maps for all the $n$ cells of $X$. By excision, $\Phi^*$ is an isomorphism for $h^*$ and $k^*$, so by naturality it suffices to show that $\mu$ is an isomorphism for $(X,A) = \coprod_\alpha (D_\alpha^n,\partial D_\alpha^n)$. The axiom for disjoint unions gives a further reduction to the case of the pair $(D^n, \partial D^n)$.
I don't think this union is disjoint - It's plausible that the boundary of the cells intersect. Isn't it? If this isn't a disjoint union, how can one fix the argument?
It seems to me that you are confused by the word "excision". Hatcher considers the two CW-pairs $\coprod_{\alpha \in I} (D_\alpha^n,\partial D_\alpha^n) = (\coprod_{\alpha \in I} D_\alpha^n,\coprod_{\alpha \in I} \partial D_\alpha^n)$ and $(X^n,X^{n-1}) $ and the map $\Phi : \coprod_{\alpha \in I} (D_\alpha^n,\partial D_\alpha^n)\to (X^n,X^{n-1})$ determined by the collection of all characteristic maps $\phi_\alpha : D^n \to X^n$, $\alpha \in I$, for the $n$-cells of $X$.
$\coprod_{\alpha \in I} (D_\alpha^n,\partial D_\alpha^n) = (D^n,\partial D^n) \times I$ is the union of the disjoint copies $(D_\alpha^n,\partial D_\alpha^n) = (D^n,\partial D^n) \times \{\alpha\}$ of $(D^n,\partial D^n)$. That is, it is the disjoint union of these spaces.
Going to quotients we obtain a commutiave diagram $\require{AMScd}$ \begin{CD} \coprod_{\alpha \in I} (D_\alpha^n,\partial D_\alpha^n) @>{\Phi}>> (X^n,X^{n-1}) \\ @V{q}VV @VV{q}V \\ (\coprod_{\alpha \in I} D_\alpha^n / \coprod_{\alpha \in I} \partial D_\alpha^n) @>>{\hat \Phi}> (X^n/X^{n-1}) \end{CD} The induced map $\hat \Phi : (\coprod_{\alpha \in I} D_\alpha^n / \coprod_{\alpha \in I} \partial D_\alpha^n) \to (X^n/X^{n-1})$ is a homeomorphism.
What Hatcher means by excision here is covered by Proposition 2.22 which is a corollary of the usual excision axiom. Applying it gives us a commutative diagram
$\require{AMScd}$ \begin{CD} H_*(\coprod_{\alpha \in I} (D_\alpha^n,\partial D_\alpha^n)) @>{\Phi_*}>> H_*(X^n,X^{n-1}) \\ @V{q_*}VV @VV{q_*}V \\ \tilde H_*(\coprod_{\alpha \in I} D_\alpha^n / \coprod_{\alpha \in I} \partial D_\alpha^n) @>>{\hat \Phi_*}> \tilde H_*(X^n/X^{n-1}) \end{CD} in which the vertical arrows are isomorphimsms. This proves that $\Phi_*$ is an isomorphism.