Let me sketch a proof of the result (which I would guess is the one Lang gives) with some motivation that I hope resonates with your question.

There is a general adjunction, if $A \to B$ is a ring homomorphism and $M$ and $N$ are $A$- and $B$-modules respectively, of the form
$Hom_A(N,M) = Hom_B(N,Hom_A(B,M)).$
This shows that if $M$ is injective over $A$, then $Hom_A(B,M)$ is injective as a $B$-module,
and also that if $N \hookrightarrow M$ is an embedding of $A$-modules, then the induced map
$N \to Hom_A(B,M)$ is also an embedding.

Thus it suffices to construct enough injectives in the category of $\mathbb Z$-modules;
the above considerations then produce enough injectives for modules over any ring.

Now $I$ is injective if and only if $Hom(\text{--},I)$ is exact, which is already a kind of Pontrjagin duality property: it gives an exact functor reversing arrows. Furthermore,
we always have a double duality map $M \to Hom(Hom(M,I))$, and another feature of Pontrjagin duality is that this map should be an isomorphism. Let's here consider the weaker property that it is injective (so we at least have a non-degenarate duality, although not necessarily perfect).

Suppose then that $I$ is some injective module over $\mathbb Z$, with the property that
the double duality map $M \to Hom(Hom(M,I),I)$ is injective for all $M$ (i.e. with the property that it gives rise to some kind of Pontrjagin duality). Then
given this $I$, if we take a free module surjecting onto $Hom(M,I),$ and then apply
$Hom(\text{--},I)$ and compose with the double duality map, we get an embedding from
$M$ into a product of copies of $I$, and hence an embedding of $M$ into an injective.

Thus constructing enough injectives is reduced to finding such $I$, and here we are
guided by the idea that we are looking for an object that establishes some kind of
(at least weak form of) Pontrjagin duality.

Now in order for the double duality to be injective, one needs that for any $m \in M$,
there is a map from $\langle m \rangle$ to $I$ that is non-zero on $m$ (since this will extend to
all of $M$ by injectivity of $I$). So we need $I$ to be injective, i.e. divisible,
and to contain torsion elements of arbitrary order. Thus we could take $I = {\mathbb Q}/{\mathbb Z}$, as in Lang, or also the circle group $S^1$. (But the latter has infinite order elements which are not necessary, so ${\mathbb Q}/{\mathbb Z}$ is more economical.
Since it also has a more algebraic/less analytic feel than the circle, this is probably why it is more commonly used in such arguments.)

Pontrjagin dual(yes, even the spelling is the same!). This is to distinguish it from the "R-dual" $M^* = \operatorname{Hom}_R(M,R)$ which comes up e.g. in the study of projective modules. $\endgroup$ – Pete L. Clark Feb 23 '11 at 14:13