This is an answer to the motivation for the definition of CW-complex. The main one is given by the following from Section $5$ of Whitehead's paper Combinatorial Homotopy I.
Let $K$ be a CW-complex.
(A) A map $f: X \to Y$ of a closed (open) subset, $ X \subset K$ in any space, $Y$, is continuous provided $f| X \cap \bar{e}$ is continuous for each cell $e \in K$.
Nowadays we might do it a bit differently and say that the topology on $K$ is defined so that this property holds. This can be done by defining the $n$ skeleton $K^n$ to be obtained from the $(n-1)$-skeleton $K^{n-1}$ by attaching $n$-cells $e^n_\lambda$ by maps $a_\lambda: S^{n-1} \to K^{n-1}$. Then define the topology on $K$ so that a map $f: K \to X$ is continuous if and only if the restriction of $f$ to $K^n \to Y$ is continuous for all $n \geqslant 0$.
This accords with the modern view that topology is "about" the category of spaces and continuous maps, and to construct these maps we often use the universal properties of pushouts and colimits (and for other examples, pullbacks and limits).
A gentle introduction to adjunction spaces and finite cell complexes is given in the book Topology and Groupoids.
Ioan James once said at Oxford that Whitehead took a year to prove the product property (H), which is now subsumed under the notion of convenient category of spaces.
His motivation can also be seen in his highly original 1941 papers in the Annals of Math, referred to in CHI. The latter paper, and CHII and Simple Homotopy Types, constitute a rewrite and extension of these essentially prewar papers.
We have inclusions $i_k : X_k \to X$. The skeleta $X_i$ have topologies, and with respect to these topologies each $X_i$ is a subspace of $X_{i+1}$.
The CW-complex $X$ is then endowed with the final topology with respect to the family $(i_k)$, i.e. with the finest topology such that all $i_k$ become continuous.
So why is it called the weak topology which in modern terms would in fact correspond to the concept of an initial topology?
It seems to me that in older literature the word "weak topology" was used as follows:
Given a set $X$ and a family $\{ X_\iota \}_{\iota \in I}$ of subsets of $X$ with each $X_\iota$ having a topology and assume that
a) $X_\iota \cap X_{\iota'}$ inherits the same subspace topology from $X_\iota$ and from $X_{\iota'}$
b) Either each $X_\iota \cap X_{\iota'}$ is open in both $X_\iota$ and $X_{\iota'}$ or each $X_\iota \cap X_{\iota'}$ is closed in both $X_\iota$ and $X_{\iota'}$
Then the weak topology on $X$ induced by $\{ X_\iota \}_{\iota \in I}$ consists of all $U \subset X$ such that each $U \cap X_\iota$ is open in $X_\iota$ (equivalently one can define closed sets $A$ by requiring $A \cap X_\iota$ closed in $X_\iota$).
See for example
Dugundji, James. "Topology Allyn and Bacon." Inc., Boston 5 (1966).
In modern language this topology would be called the final topology induced by the inclusions $X_\iota \to X$.
Added:
The first occurrence of the phrase "weak topology" I could find was in
Whitehead, John HC. "Combinatorial homotopy. I." Bulletin of the American Mathematical Society 55.3 (1949): 213-245.
In this paper the concept of "CW-complex" was introduced. However, it goes back to
Whitehead, John Henry Constantine. "Simplicial Spaces, Nuclei and m‐Groups." Proceedings of the London mathematical society 2.1 (1939): 243-327.
On p.316 the concept of a "topological polyhedron" was introduced. Such a space is given the weak topology with respect to its closed cells (although the word "weak topology" was not used).
In a more general context weak topologies have been considered for example in
Morita, Kiiti. "On spaces having the weak topology with respect to closed coverings." Proceedings of the Japan Academy 29.10 (1953): 537-543.
Cohen, D. E. "Spaces with weak topology." The Quarterly Journal of Mathematics 5.1 (1954): 77-80.
See also p.44 of
Hart, Klaas Pieter, Jun-iti Nagata, and Jerry E. Vaughan. Encyclopedia of general topology. Elsevier, 2003.
All these sources do not really clarify why the word "weak" was used. Intuitively one would expect that a weak topology on a set has fewer open sets (i.e. is coarser) than other possible topologies. For example, on polyhedra the weak topology (= CW topology) is in general finer then the metric topology and thus one could regard it as stronger than the metric topology.
The usage seems to have again historical reasons. In
Arens, Richard F. "A topology for spaces of transformations." Annals of Mathematics (1946): 480-495.
I found the following in section 3 "Comparison of topologies":
Suppose that $t$ and $t^*$ are two topologies for the same class of elements. If the open sets of $t$ are also open in $t^*$, we shall write $t \subset t^*$, and say that $t$ is stronger than $t^*$, and $t^*$ is weaker than $t$.
We see that weaker is the same as finer is modern language. However, there is a footnote saying
In our use of "strong" and "weak", we concur with Alexandroff and Hopf (1, p. 62), but not all writers. "Stronger", above, means roughly "more limit points".
The quoted book by Alexandroff and Hopf (in German) is from 1935. It is still available:
Alexandroff, Paul, and Heinz Hopf. Topologie I: Erster Band. Grundbegriffe der Mengentheoretischen Topologie Topologie der Komplexe· Topologische Invarianzsätze und Anschliessende Begriffsbildungen· Verschlingungen im n-Dimensionalen Euklidischen Raum Stetige Abbildungen von Polyedern. Springer-Verlag, 2013.
In this book the concept of a topological space is introduced as a set $X$ with (in modern language) a closure operator for subsets $M \subset X$ ($M \mapsto \overline{M}$). The authors use the word topological assignment to denote a closure operator.
This is no longer the standard approach, but it is equivalent to defining a topological space to be a set with a topology (system of open subsets).
For those who are historically interested let me give a translation of the relevant definition on p.62:
If two topological spaces $X_1, X_2$ have the same underlying set, and if the identity map $X_1 \to X_2$ is continuous, we say that the topological assignment on $X_1$ is at most as strong as that on $X_2$.
A footnote says
This denotation is justified: If the topological assignment on $X_1$ is at most as strong as that on $X_2$, then this means: If $p$ is a limit point of $M$ in $X_1$, then it is certainly a limit point of $M$ in $X_2$; but $p$ may be a limit point of $M$ in $X_2$ without being a limit point of $M$ in $X_1$. Thus the topological assignment on $X_2$ is a strengthening of the topological assignment on $X_1$.
Although the words "stronger" and "weaker" are not explicitiy used here, it is obvious that if the topology (= topological assignment) on $X_1$ is at most as strong as that on $X_2$, then the topology on $X_2$ is regarded as at least as strong as that of $X_1$, in other words as stronger. Thus the topology on $X_1$ is weaker than that on $X_2$. In terms of closure operators it means that $\overline{M}^{1} \subset \overline{M}^{2}$ for all subsets $M$. Whether the second closure operator should then be regarded as a strenghtening of the first is a philosophical question. But certainly one can argue that it "stronger absorbs points" than the first.
Update:
The above interpretation of a stronger topology also occurs in
Barratt, M. G., and John Milnor. "An example of anomalous singular homology." Proceedings of the American Mathematical Society 13.2 (1962): 293-297.
In this paper the concept of a bouquet with the strong topology is introduced. The authors introduce the so-called strong topology on the one-point union of a family of spaces with basepoint and write "The resulting topology is stronger than the usual weak topology". Here the weak topology is again the final topology induced by the injections of the spaces into the one-point union. In modern language one would say that the strong topology on the one-point union is coarser than the weak topology.
Best Answer
A CW-complex $X$ goes along with a cell-decomposition $\mathcal E$ and is coherent with the collection of closed cells $\{\overline e\mid e\in\mathcal E\}$.
That means that a set $A\subseteq X$ is open if and only if $A\cap\overline{e}$ is open in $\overline e$ for every $e\in\mathcal E$. So "yes" to your first question.
Actually this statement is equivalent with the one that states that $A\subseteq X$ is closed if and only if $A\cap\overline{e}$ is closed in $\overline e$ for every $e\in\mathcal E$.
You are correct in thinking that an open cell $e$ is not necessarily open in $X$. For instance we can think of $S^1$ as a union $e^0\cup e^1$ where $e^0=\{0\}$ and $e^1$ are disjoint open cells. It is evident that $e^0=\{0\}$ is not an open subset of $S^1$.
Defining $\mathbb{E}^{n}=\left\{ x\in\mathbb{R}^{n}\mid\left\Vert x\right\Vert <1\right\} $ and $\mathbb{D}^{n}=\left\{ x\in\mathbb{R}^{n}\mid\left\Vert x\right\Vert \leq1\right\} $ as open and closed $n$-disk an $n$-cell is a space homeomorphic to $\mathbb{E}^{n}$ and inherits the label "open" from $\mathbb{E}^{n}$.