= Solution
Let $a,b\in H^2(S^2\times S^2;\mathbb Z)$ be the two factor classes, normalized by $a^2=b^2=0$ and $\langle ab,[X]\rangle=1$. Choose the <complex line bundle> $L$ with $c_1(L)=a+b$, for example the tensor product of pullbacks of degree-one <complex line bundles> from the two factors. Then
$$
c_1(L)^2[X]=2.
$$
If $L$ had an <ASD connection>, the real closed form $\alpha=iF/(2\pi)$ would represent this class and satisfy $*\alpha=-\alpha$. Hence
$$
2=\int_X\alpha\wedge\alpha=-\int_X|\alpha|^2\,d\mathrm{vol}_g\leq0,
$$
a contradiction. \b[This <complex line bundle> admits no ASD connection for any metric of the given product orientation.] For the reverse product orientation, take $c_1(L)=a-b$ instead, which then has square $+2$. This is the <positive-square obstruction to ASD line connections>.
The suggested rank-two construction gives the same obstruction. The split bundle $E=L\oplus L^*$ has trivial determinant and
$$
c(E)=(1+c_1(L))(1-c_1(L)),\qquad c_2(E)[X]=-c_1(L)^2[X]=-2.
$$
An <ASD connection> on $L$ would induce an <ASD connection> on $E$, but its <instanton number> would be $\|F_E\|_2^2/(8\pi^2)\geq0$, contradicting $c_2(E)[X]=-2$.
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