We shall assume that the $X_i$'s are independent. The problem can be restated as follows: show that for some $h\in(0,1)$, all natural $n$, and all subsets $J$ and $K$ of the set $[n]:=\{1,\dots,n\}$ 
we have 
\begin{equation*}
S:=\sum_{x,y}|P(X_J=x,X_K=y)-P(X_J=x+1,X_K=y+1)|\le2-h, \tag{1}	
\end{equation*}
where $X_J:=\sum_{i\in J}X_i$ and the sum in (1) is over all all integers $x,y$. Write 
\begin{equation*}
	S\le T+U, \tag{2}
\end{equation*}
where 
\begin{equation*}
T:=\sum_{x,y}|P(X_J=x,X_K=y)-P(X_J=x+1,X_K=y)|,
\end{equation*}
\begin{align*}
U&:=\sum_{x,y}|P(X_J=x+1,X_K=y)-P(X_J=x+1,X_K=y+1)| \\  
&=\sum_{x,y}|P(X_J=x,X_K=y)-P(X_J=x,X_K=y+1)|. 	
\end{align*} 

By the independence of the $X_i$'s, 
\begin{equation*}
	P(X_J=x,X_K=y)=\sum_z P(X_{J\cap K}=z)P(X_{J\setminus K}=x-z)P(X_{K\setminus J}=y-z). 
\end{equation*}
Hence, 
\begin{align*}
T&\le\sum_zP(X_{J\cap K}=z)\,\sum_y P(X_{K\setminus J}=y-z) \\ 
&\times\sum_x|P(X_{J\setminus K}=x-z)-P(X_{J\setminus K}=x+1-z)| \\ 
&=\sum_x|P(X_{J\setminus K}=x)-P(X_{J\setminus K}=x+1)|=:D_{|J\setminus K|}, 
\end{align*}
where $|\cdot|$ denotes the cardinality. 
Similarly, $U\le D_{|K\setminus J|}$, so that, by (2) 
\begin{equation*}
	S\le D_{|J\setminus K|}+D_{|K\setminus J|}. \tag{3}
\end{equation*}
Note that $D_0=1$ and, by [this answer][1], for $k\ge1$ we have 
\begin{align*}
	D_k=\frac1{2^k}\,\Big(2\binom k{m+1}-1\Big)\le\frac34,  
\end{align*}
where $m:=\lfloor (k-1)/2\rfloor$. So, by (3), 
\begin{equation*}
	S\le 1+3/4=7/4 
\end{equation*}
if $J\ne K$. 

In the remaining case when $J=K$, 
\begin{equation*}
S=\sum_x|P(X_J=x)-P(X_J=x+1)|=D_{|J|}\le1. 	
\end{equation*}

Thus, in all cases (1) holds with $h=2-\max(7/4,1)=1/4>0$, as desired. 

[1]: https://mathoverflow.net/a/307506/36721