So I am trying to understand the topology on cw complexes. I know that the $n+1$ skeleton of X is formed as
$$X^{n+1} = \{ X^n \sqcup D_{i}^{n+1} \}/ \sim$$
where $\sim$ denotes $s \sim f_i^{n+1}(s)$ where $s \in \partial D_{i}^{n+1}$ and $f_{i}^{n+1}: \partial D_i^{n+1} \to X^n$ are the attaching maps.
Now I know X is the union of all the n-sletelons.
Now I want to know is each cell closed? I was thinking with the weak topology a set $A$ is closed in X if $A \cap X^n$ is closed in each $X^n$. So an n-cell must be closed in each skeleton since $X$ is formed by taking a closed n-cell and attaching it along the boundary.
So each $n$-skeleton $X^n$ is closed? Can I extend this idea to say that $X^n$ is compact because $X^0$ is finite and each n-cell is compact in $\mathbb{R}^n$ (by H.B), so that makes $X^0 \sqcup D_i^1$ compact so then $X^1$is compact since the quotient map is continuous?
You are right, each $n$-skeleton $X^n$ is closed in $X$. But $X^0$ is not required to be finite and the attaching construction $X^n \mapsto X^{n+1}$ may involve infinitely many $D_i^{n+1}$. Therefore you cannot conclude that the $X^n$ are compact.
A simple example is $X = \mathbb R$ with $X^0 = \mathbb Z$ (which has the discrete topology). Then $X^1 = \mathbb R$ is obtained by attaching the cells $D^1_i = D^1$, $i \in \mathbb Z$, via the maps $f^1_i : S^0_i \to X^0, f_i^1(-1) = i - 1, f_i^1(1) = i$.