Discrete elements of the coset ring are generated by discrete cosets

Put a finite Boolean expression for $E$ in disjunctive normal form. Each nonempty cell has the form $A\setminus {\bigcup }_{j=1}^m{B}_j$, where $A$ and the $B_j$ are cosets: intersecting positive cosets gives a coset or the empty set, and a cell with no positive literal uses $A={\mathbb{R}}^d$. Thus $E$ is a finite union of such cells.

We first prove the cover claim: if a non-discrete coset $A=x+H$ is covered by a locally finite set $F\subseteq A$ and finitely many cosets $B_j$ of subgroups of $H$, each disjoint from $F$, then $F=\mathrm{\varnothing }$. Induct on the number $k$ of distinct subgroups defining the $B_j$. If a subgroup $K$ is locally finite in ${\mathbb{R}}^d$, a compact neighborhood of $0$ meets $K$ in finitely many points, so shrinking it to exclude the nonzero points gives a neighborhood $U$ with $U\cap K=\{0\}$. Choose a symmetric neighborhood $V$ with $V-V\subseteq U$. Every compact set has a finite cover by translates of $V$, and each translate meets any coset of $K$ in at most one point. Hence every discrete subgroup coset is locally finite. If every subgroup among the $B_j$ were discrete, then their finite union with $F$ would be locally finite, forcing $A$ to be discrete. Therefore at least one defining subgroup, say $K_1$, is non-discrete. If $k=1$ and $y\in F$, then $y+K_1\subseteq A$ cannot meet any $B_j$: a $K_1$-coset meeting $y+K_1$ equals it and would contain $y$, contrary to $B_j\cap F=\mathrm{\varnothing }$. Thus $y+K_1\subseteq F$, contradicting local finiteness of $F$ because $K_1$ is non-discrete. For $k>1$, the same argument shows that $y+K_1$ is covered only by $F$ and the cosets whose defining subgroups differ from $K_1$. Translate by $-y$ and intersect with $K_1$. Each nonempty intersection with such a coset is a coset in $K_1$ of the proper subgroup $K_1\cap K_j$; it is proper because otherwise that coset would contain $y$. The set $F^{\mathrm{\prime }}=(F-y)\cap K_1$ is locally finite and contains $0$, and is disjoint from all these intersection cosets. The induction hypothesis inside the non-discrete group $K_1$, with fewer than $k$ distinct subgroups, says $F^{\mathrm{\prime }}=\mathrm{\varnothing }$, a contradiction. This proves the cover claim for every $k$.

Let $C=A\setminus {\bigcup }_j{B}_j$ be a cell in step 1.1 and suppose $C$ is nonempty and discrete. If $A$ were non-discrete, then $A=C\cup {\bigcup }_j(A\cap B_j)$ would be a cover of the type in step 1.2: $C$ is locally finite and each $A\cap B_j$ is a coset disjoint from $C$. Step 1.2 would force $C=\mathrm{\varnothing }$, contrary to the supposition. Thus $A$ is a discrete coset. Every $A\cap B_j$ is then also a discrete coset, so $C$ belongs to the ring generated by discrete cosets. Each cell in the finite union for $E$ is a subset of the locally finite set $E$, hence is locally finite; applying this argument to every nonempty cell and taking their finite union proves the statement.