Sampling distribution from two independent Bernoulli populations

Put
$a=\frac{\sqrt{\theta_1(1-\theta_1)}}{\sqrt{n_1}}$, $b=\frac{\sqrt{\theta_2(1-\theta_2)}}{\sqrt{n_2}}$,
$A=(\bar{X}_1-\theta_1)/a$,
$B=(\bar{X}_2-\theta_2)/b$. We have
$A\to_d N(0,1),\ B\to_d N(0,1)$.
In terms of characteristic functions it means
$$\phi_A(t)\equiv {\bf E}e^{itA}\to e^{-t^2/2},\ \phi_B(t)\to e^{-t^2/2}.$$
We want to prove that
$$D:=\frac{a}{\sqrt{a^2+b^2}}A-\frac{b}{\sqrt{a^2+b^2}}B\to_d N(0,1)$$

Since $A$ and $B$ are independent,
$$\phi_D(t)=\phi_A\left(\frac{a}{\sqrt{a^2+b^2}}t\right)\phi_B\left(-\frac{b}{\sqrt{a^2+b^2}}t\right)\to e^{-t^2/2},$$
as we wish it to be.

This proof is incomplete. Here we need some estimates for uniform convergence of characteristic functions. However in the case under consideration we can do explicit calculations. Put $p=\theta_1,\ m=n_1$.
$$\begin{align} \phi_{X_{1,1}}(t) &= 1+p(e^{it}-1), \\ \phi_{\bar X_{1}}(t) &= (1+p(e^{it/m}-1))^m, \\ \phi_{\bar X_{1}-\theta_1}(t) &= (1+p(e^{it/m}-1))^m e^{-ipt}, \\ \phi_{A}(t) &= (1+p(e^{it/\sqrt{mp(1-p)}}-1))^m e^{-ipt\sqrt{m}/\sqrt{p(1-p)}} \\[5pt] &= \left( \left(1+p(e^{it/\sqrt{mp(1-p)}}-1)\right)e^{-ipt/\sqrt{mp(1-p)}}\right)^m \\[5pt] &=\left( 1-\frac{t^2}{2m}+O(t^3m^{-3/2}) \right)^m \end{align}$$
as $t^3m^{-3/2}\to 0$. Thus, for a fixed $t$,
$$\phi_D(t)=\left( 1-\frac{a^2t^2}{2(a^2+b^2)n_1}+O(n_1^{-3/2}) \right)^{n_1} \left( 1-\frac{b^2t^2}{2(a^2+b^2)n_2}+O(n_2^{-3/2}) \right)^{n_2} \to e^{-t^2/2}$$
(even if $a\to 0$ or $b\to 0$), since $\left|e^{-y}-(1-y/m)^m\right|\le {y^2}/{2m}\ $ when $\ y/m<1/2$ (see https://math.stackexchange.com/questions/2566469/uniform-bounds-for-1-y-nn-exp-y/ ).

Note that similar calculations may be done for arbitrary (not necessarily Bernoulli) distributions with finite second moments, using the expansion of characteristic function in terms of the first two moments.