closed cells form a covering of the CW complex. A CW complex $X$ is the union of open cells $\{e^i_\alpha\}$ and these open cells are disjoint subsets of 
$X$, so these open cells form a partition of $X$.
Now if we take the closed cells $\{\bar e^i_\alpha\}$, then if the number of cells is finite we can also write $X$ as a union of closed cells
$$X=\bigcup_{i,\alpha}\bar e^i_\alpha$$ where the $\bar e^i_\alpha$ are not necessarily disjoint sets. In sum, the open cells form a partition of $X$ whereas the closed cells form a covering of $X$.
When the number of cells is infinite we don't have a covering of $X$ by closed cells, we only have that the union of closed cells is contained in $X$.
Is this correct?
 A: As Lord Shark the Unknown pointed out, you always have $X=\bigcup_{i,\alpha}\bar e^i_\alpha$ (simply because $e^i_\alpha \subset \bar e^i_\alpha$).
The set of open cells on its own does not determine the topology of the whole space $X$. The open cells are topologically fairly trivial as they are all homeomorphic to standard open balls in some Euclidean space, and elementary examples that non-homeomorphic spaces may be partitioned into "the same" collection of open cells (consider $X= [0,1]$ and $X' = S^1 \cup \{ 0 \}$ which both can be decomposed in two $0$-cells and one $1$-cell).
Thus we need to know how the open cells are pieced together, and that is why the closed cells enter the arena: Only the intersections $\bar e^i_\alpha \cap e^j_\beta = (\bar e^i_\alpha \setminus e^i_\alpha) \cap e^j_\beta$ are able to provide information about the topology of $X$, and for this purpose the characteristic maps $\phi^i_\alpha : (D^i,\mathring{D}^i) \to (\bar e^i_\alpha, e^i_\alpha)$ are needed.
In fact, there are various equivalent approaches to define CW complexes. See 
Show that any cell complex is homotopy equivalent to a CW complex and consult a textbook on that material.
