= Solution
We use two precise facts about the <formal group of an elliptic curve>. At a <prime> $p$ of <good reduction>, its <kernel of reduction of an elliptic curve> is identified by the <uniformizer> $t=-x/y$ with the group $\widehat E(p\mathbb Z_p)$. For odd $p$ this group is torsion-free. To see the second fact, the integral <invariant differential of a formal group law> has the form $(1+\sum_{j\geq1}c_jT^j)dT$, with $c_j\in\mathbb Z_p$. Integrating constructs the <formal logarithm>
$$
\log_F(T)=T+\sum_{j\geq2}d_jT^j/j,\qquad d_j\in\mathbb Z_p.
$$
It is a <group homomorphism> to the <additive group>. For $0\ne t\in p\mathbb Z_p$ and $j\geq2$,
$$
v_p(d_jt^j/j)\geq jv_p(t)-v_p(j)>v_p(t)
$$
when $p$ is odd. Hence the series converges and $v_p(\log_F(t))=v_p(t)$, so it is injective. The target has no nonzero torsion. Therefore reduction is injective on the entire rational <torsion subgroup> at an odd <prime> of <good reduction>, including its $p$-primary part.
For the present <elliptic curve>, $\Delta=64D^6$, so every odd $p\nmid D$ is a <prime> of <good reduction>. If $p\equiv3\pmod4$, the <Legendre symbol> of $-1$ is $-1$. The values $x$ and $-x$ cancel in the sum of the <Legendre symbols> of $x^3-D^2x$, yielding
$$
\#E(\mathbb F_p)=p+1.
$$
The rational <torsion subgroup> injects into each of these groups, so it is finite and its order $M$ divides every such $p+1$.
For any odd <prime> $\ell$, the <Dirichlet theorem on primes in arithmetic progressions> supplies infinitely many $p$ with $p\equiv3\pmod4$ and $p\equiv1\pmod\ell$. Discard the finitely many dividing $D$. Since $\ell\nmid p+1$, it cannot divide $M$. Similarly choose $p\equiv3\pmod8$, again avoiding $D$; then $v_2(p+1)=2$, so $M\mid4$.
There are already four rational <2-torsion> points,
$$
O,\quad(0,0),\quad(D,0),\quad(-D,0),
$$
which are distinct because a squarefree integer $D$ is nonzero. Thus
$$
\boxed{E(\mathbb Q)_{\mathrm{tors}}=E(\mathbb Q)[2]\cong(\mathbb Z/2\mathbb Z)^2.}
$$
\b[Its order is four.] In fact the argument works for every nonzero integer $D$; squarefreeness is not needed for this torsion conclusion.
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