Bernoulli root laws are uniformly close in Levy–Prokhorov distance

If $p=q$, the laws coincide. Suppose $p\le q$. For each $i$, give $(B_i,C_i)\in \{0,1{\}}^2$ probabilities $p$ at $(1,1)$, $q-p$ at $(0,1)$, $1-q$ at $(0,0)$, and zero at $(1,0)$, and take $n$ independent copies of this finite pair. Then $B_i\sim \mathrm{Bernoulli}(p)$, $C_i\sim \mathrm{Bernoulli}(q)$, and $D_i:=C_i-B_i\sim \mathrm{Bernoulli}(q-p)$; the pairs are independent across $i$. Thus $S_{n,q}-S_{n,p}=n^{-1\mathrm{/}2}{\sum }_i(D_i-(q-p))$, whose mean is zero and whose variance, by [F2], is $(q-p)(1-q+p)\le q-p$. If $q<p$, interchange the two parameters and change the sign of the difference; its variance is $\mathrm{\mid }p-q\mathrm{\mid }(1-\mathrm{\mid }p-q\mathrm{\mid })\le \mathrm{\mid }p-q\mathrm{\mid }$.

Put $r=\mathrm{\mid }p-q\mathrm{\mid }$ and $\epsilon =r^{1\mathrm{/}3}$ for $r>0$. By [F3] and step 1.1, the coupling just constructed satisfies $\mathbb{P}(\mathrm{\mid }{S}_{n,p}-S_{n,q}\mathrm{\mid }>\epsilon )\le r\mathrm{/}{\epsilon }^2=\epsilon$. For any closed set $A$, the event $\{S_{n,p}\in A,\mathrm{\mid }{S}_{n,p}-S_{n,q}\mathrm{\mid }\le \epsilon \}$ is contained in $\{S_{n,q}\in A^{[\epsilon ]}\}$, so ${\mu }_{n,p}(A)\le {\mu }_{n,q}(A^{[\epsilon ]})+\epsilon$. Interchanging $p$ and $q$ gives the reverse inequality. Hence $\epsilon$ is admissible in [F4], so the infimum defining $\pi$ is at most $\epsilon$. For $r=0$ the laws coincide and their distance is zero by [F4]. This proves the bound for every endpoint case as well.