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Use integral homology groups. The sphere is $A=S^3$; each disk meets $A$ in its boundary circle, while $B\cap C=\{0\}$. In particular, the two disks also meet away from the sphere. These spaces have compatible triangulations, so the Mayer–Vietoris sequence applies to their unions as subcomplexes; equivalently use small regular open neighborhoods with the same homotopy types.

First let $Y=A\cup B$. Its intersection is $A\cap B=S^1$, and $B$ is contractible. The Mayer–Vietoris sequence givesFor example, the degree-two portion is $0\to H_2(Y)\to H_1(S^1)\to 0$, and the map from $H_0(S^1)$ into $H_0(A)\oplus H_0(B)$ is injective.

Now $Y\cap C=\partial C\bigsqcup \{0\}\simeq S^1\bigsqcup \{\text{point}\}$. It has $H_1=\mathbb{Z}$ and $H_0={\mathbb{Z}}^2$. Exactness gives $H_3(X)=H_3(Y)=\mathbb{Z}$ andThis sequence splits: lift a generator of the quotient $\mathbb{Z}$ and extend its lift by integer multiples. Hence $H_2(X)={\mathbb{Z}}^2$. In degree one the sequence becomesThe middle map has kernel generated by $(1,-1)$ and cokernel $\mathbb{Z}$. There are no higher groups, either by the same sequence or by the three-dimensional triangulation. Thus the homology of two disks meeting inside a sphere givesThe first-homology generator records a path from the sphere to the common interior point through one disk and back through the other; omitting that interior intersection would incorrectly remove this generator.