Let Fj be the closure of Ej. Other points from Ri may be present in the boundary of Fj. However, since a finite number of cells contain the boundary, the intersection of Fj and Ri is finite.
Since S is hausdorff, each point is closed, and Ej∩Ri is closed. Thus each Ri is closed in S.
Let Oi be the complement of Ri. Note that Oi forms a cover for S, and for T. Any finite subcover would exclude some point pn, which contradicts T a compact set. Therefore T intersects a finite number of cells.
This collection of cells, call it U, intersects a closed cell in the entire closed cell or nothing. The intersection is always closed, hence U is closed. We want to prove U is a cw complex, that is, a subcomplex of S.
Each characteristic function from a closed ball into the cells of U remains valid. The image of the boundary is covered by finitely many cells of S at lower dimensions, and all these cells are folded into U.
If a subset V is closed in U it is closed in S, and its intersection with any closed cell of S, or of U, remains closed.
This reasoning can be reversed. Let V intersect every closed cell of U in a closed set. Since U is closed, this takes care of all the cells of U. Let E be a cell not contained in U, which means E lies outside of U, since cells are disjoint. If V intersects the boundary of E, it intersects lower dimensional cells (containing the boundary) in closed sets. Combine these (finitely many) intersections together, and the result is still closed. Thus V intersects every closed cell of S in a closed set, and V is closed in S. With V entirely inside U, V remains closed in the subspace topology of U. Therefore U is a cw subcomplex of S.