There are several ways - for example induction on $n$ should work. We can also use the stacked basis theorem, if so desired.
But let's do the following. Select a basis ${\mathcal B}_A =\{a_1,a_2,\ldots,a_n\}$ of $A$.
Let us similarly select a basis ${\mathcal B}_B=\{b_i,b_2,\ldots, b_n\}$ of $B$. As $B\subseteq A$ we get a matrix
$M=(m_{ij})$, with integer entries, uniquely determined by
$$
b_i=\sum_j m_{ij}a_i.
$$
I want to show that the matrix $M$ is non-singular.
Let $V$ be the vector space over $\mathbf{Q}$ with basis ${\mathcal B}_A$. Then ${\mathcal B}_A$
and ${\mathcal B}_B$ are both linearly independent over $\mathbf{Z}$ in $V$, because they were so in $A$. Thus they are both vector space bases of $V$, and $M$ is the change of bases matrix that is then non-singular.
The inverse matrix $M^{-1}$ then has the property that $(\det M)M^{-1}$ also has integer entries (this follows immediately from Cramer's rule). This means that (working inside $V$ again) the basis elements $a_i$ have the property that $(\det M)a_i$ is a linear combination of vectors $b_j$ with integer coefficients. Therefore $a_i\in(\det M)B$. As this holds for all $a_i$, we have shown that
$$
(\det M)A\subseteq B.
$$
The claim follows from this: $A/B$ is a finitely generated abelian group with exponent that is a factor of $|\det M|$, so it must be finite.
A more careful study (using stacked bases theorem) reveals that in fact we have
$$
[A:B]=|\det M|.
$$