Hatcher describes how CW-complexes are constructed:
Start with a set $X^0$ having the discrete topology.
Construct inductively skeleta $X^n$ by attaching $n$-cells to $X^{n-1}$. Here $X^{n-1}$ already has a topology and $X^n$ is defined as a suitable quotient space of $X^{n-1}\bigsqcup_\alpha \mathbb{D}_\alpha^n$.
This construction produces a sequence of spaces $X^0, X^1, X^2,\ldots$ such that $X^{n-1}$ is a subspace of $X^n$.
If this process stops at some finite $N$, then we have a topology on $X = X^N$. In that case trivially $A \subset X$ is open iff $A \cap X^n$ is open in $X^n$ for all $n$. Just note that $X^n = X$ for $n \ge N$.
If the attaching process continues ad infinitum, then we have a topology on each skeleton $X^n$, and these toplogies are compatible in the sense that $X^{n-1}$ is a subspace of $X^n$ for each $n$. Howewer, we do not have a topology on $X = \bigcup_{n=0}^\infty X^n$. That is why we give the space $X$ the weak topology defined as in (3). This definition is relevant only in case that the attaching process continues ad infinitum.
Edited on request:
Suppose we have an ascending sequence of topological spaces $X^0, X^1, X^2,\ldots$ (ascending means that for all $n > 0$ we have $X^{n-1} \subset X^n$ and that the topology on $X^{n-1}$ is the subspace topology inherited from $X^n$). Then let $X = \bigcup_{n=0}^\infty X^n$. Which topology do we introduce on $X$ in order that all subspaces $X^n$ receive their original topology? In general there are many ways to do that, but the standard approach is to define $A \subset X$ open in $X$ iff $A \cap X^n$ is open in $X^n$ for all $n$. This is the final topology with respect to the system $\{X^n\}$. In the context of CW-complexes it is also denoted as the weak topology. This has historical reasons. See Confusion about topology on CW complex: weak or final?
Let us prove that if $X$ is endowed with the final topology, then all subspaces $X^n \subset X$ have their original toplogy:
Let $U \subset X^n$ be open in the subspace toplogy. Then $U = A \cap X^n$ with some open $A \subset X$. But by definition of the final topology $A \cap X^n$ is open in $X^n$ with its original toplogy.
Let $U = U_n \subset X^n$ be open in the original topology. Using the fact that each $X^{k-1}$ is a subspace of $X^k$, we can recursively construct open $U_k \subset X^k$, $k \ge n$, such that $U_{k+1} \cap X^k = U_k$. Moreover, for $k < n$ define $U_k = U_n \cap X^k$ which is open in $X^k$. By construction we have $U_k \cap X^m = U_k$ if $k \le m$ and $U_k \cap X^m = U_m$ if $k > m$. Then $A = \bigcup_{k=0}^\infty U_k$ is open in $X$: In fact, $A \cap X_m = (\bigcup_{k=0}^\infty U_k) \cap X_m = \bigcup_{k=0}^\infty U_k \cap X_m = \bigcup_{k=0}^m U_k \cup \bigcup_{k=m+1}^\infty U_m = U_m$ which is open in $X^m$. This also shows $A \cap X^n = U_n = U$, i.e $U$ is open in the subspace topology.
If the sequence $X^0, X^1, X^2,\ldots$ stabilizes, i.e. if we have $X = X^N$ for some $N$ (which is the same as $X^n = X^N$ for $n \ge N$), then there is no need to introduce a new topology on $X$ since $X^N$ already has one. The above proof shows nevertheless that $A \subset X$ is open iff $A \cap X^n$ is open in $X^n$ for all $n$. However, in that case it is trivial: If $A \cap X^n$ is open in $X^n$ for all, then $A = A \cap X = A \cap X^N$ is open in $X = X^N$. Conversely, if $A \subset X^N$ is open, then $A \cap X^n$ is open in $X^n$ for all $n \le N$ because $X^n$ is a subspace of $X^N$ and open in $X^n$ for all $n > N$ because $X^n = X^N$ and $A \cap X^n = A$.
Note that $e^n_{\alpha}$ is an open $n$-disk, and $D^n_{\alpha}$ is a closed $n$-disk, so $D^n_{\alpha} = e^n_{\alpha}\sqcup\partial D^n_{\alpha}$. The identification $\phi_{\alpha} : \partial D^n_{\alpha} \to X^{n-1}$ only identifies points of $\partial D^n_{\alpha}$ with points of $X^{n-1}$, the points of $e^n_{\alpha}$ do not get identified. So, as sets, we have
\begin{align*}
X^n &= (X^{n-1}\sqcup_{\alpha} D^n_{\alpha})/(x\sim \phi_{\alpha}(x))\\
&= (X^{n-1}\sqcup_{\alpha}\partial D^n_{\alpha}\sqcup_{\alpha}e^n_{\alpha})/(x\sim\phi_{\alpha}(x))\\
&= (X^{n-1}\sqcup_{\alpha}\partial D^n_{\alpha})/(x\sim\phi_{\alpha}(x))\sqcup_{\alpha} e^n_{\alpha}\\
&= X^{n-1}\sqcup_{\alpha} e^n_{\alpha}.
\end{align*}
Best Answer
The technically "clean" way is to take the colimit of the sequence of spaces.
In fact, the process of attaching cells is a special of the adjunction space construction: Given spaces $X,Y$ and a map $f : A \to Y$ defined on a subspaces $A \subset X$, we define the adjunction space $X \cup_f Y$ as the quotient space of the disjoint union $X + Y$ obtained by identifying each $a \in A$ with its image $f(a) \in Y$. This yields a canonical embedding $i : Y \to X \cup_f Y$.
But is $Y$ a genuine subspace of $X \cup_f Y$? This is a somewhat philosophical question and I think the answer is irrelevant. The purpose of the adjunction space is best explained by its universal property as a pushout; see here.
Anyway, the answer depends on the specific construction of $X \cup_f Y$. On the level of sets the standard construction of $X \cup_f Y$ produces a set of equivalence classes in $X + Y$, i.e. a subset of the power set of $X + Y$ which never contains $Y$ as a genuine subset (whatever the concrete definition of $X + Y$ may be). But with some work one can also give an alternative construction: Find a bijection $\beta : X \to X'$ to a set $X'$ such that $X' \cap Y = \emptyset$, then take $X \cup_f Y = (X' \setminus \beta(A)) \cup Y$ and give it the the appropriate topology. Then $Y \subset X \cup_f Y$.
Let us come back to CW complexes. $X^n$ is obtained as an adjunction space from $X^{n-1}$ by attaching a collection of $n$-cells. We certainly get a canonical embedding $i_n : X^{n-1} \to X^n$; but whether $X^{n-1}$ is a genuine subspace of $X^n$ depends on the chosen construction. Yes, usually one writes $ X^{n-1} \subset X^n$, but some caution should be taken.
I think the direct limit approach makes it conceptually clearer why we topologize infinite-dimensional CW-complexes by giving them the weak topology; again there is a universal property.
Note that if we have an ascending sequence of spaces $X^0 \subset X^1 \subset X^2 \subset \ldots $, then the union $\bigcup_n X^n$ can be given the weak topology as in 3. and the resulting space has the universal property of the direct limit.