I am confused about a step in the following proof:
Proposition: Every $n$-cell of an $n$-dimensional CW complex $X$ is open.
Proof: Suppose $e_0$ is an open $n$-cell of $X$, $e$ is any other (i.e. $e \neq e_0$) cell of $X$, then $e_0 \cap e = \emptyset$, so $e_0 \cap \bar e$ is contained in $\bar e\setminus e$, which in turn is contained in a union of finitely many cells of dimension less than $n$. Since $e_0$ has dimension $n$, it follows that $e_0 \cap \bar e = \emptyset$.
Why must $\bar e \setminus e$ contained in the union of finitely many cells of dimension less than $n$? It seems intuitive that the boundary of an n-cell must has dimension less than its dimension, however I cannot figure out a way to prove this.
(I am using the definition of CW complex from page 132 of Lee's Introduction to Topological Manifolds.)