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.)
This is part of the definition of a CW-complex. Specifically, following Lee's definition, the definition of a cell decomposition of a space requires that the characteristic map $\Phi:D\to X$ of an $m$-cell maps $\partial D$ to the union of all the cells of dimension strictly less than $m$. In your case, this means that since the dimension of $e$ is at most $n$, $\bar{e}\setminus e$ (which is equal to the image of $\partial D$ under the characteristic map of $e$) is contained in the union of all the cells of dimension strictly less than $n$. The fact that you need only finitely many cells of smaller dimension to cover $\bar{e}\setminus e$ follows from the closure-finite axiom for CW complexes.