= Solution
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> gives
$$
H_3(Y)=\mathbb Z,\qquad H_2(Y)=\mathbb Z,\qquad H_1(Y)=0,\qquad H_0(Y)=\mathbb Z.
$$
For example, the degree-two portion is $0\to H_2(Y)\to H_1(S^1)\to0$, and the map from $H_0(S^1)$ into $H_0(A)\oplus H_0(B)$ is injective.
Now $Y\cap C=\partial C\sqcup\{0\}\simeq S^1\sqcup\{\text{point}\}$. It has $H_1=\mathbb Z$ and $H_0=\mathbb Z^2$. Exactness gives $H_3(X)=H_3(Y)=\mathbb Z$ and
$$
0\longrightarrow\mathbb Z\longrightarrow H_2(X)\longrightarrow\mathbb Z\longrightarrow0.
$$
This <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> becomes
$$
0\longrightarrow H_1(X)\longrightarrow\mathbb Z^2
\xrightarrow{(m,n)\mapsto(m+n,-m-n)}\mathbb Z^2
\longrightarrow H_0(X)\longrightarrow0.
$$
The 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> gives
$$
\boxed{H_j(X;\mathbb Z)=\begin{cases}\mathbb Z,&j=0,1,3,\\\mathbb Z^2,&j=2,\\0,&j\ge4.\end{cases}}
$$
The 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.
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