With regard to your third question, while I agree with the comments that it's neither necessary nor beneficial to study category theory as a precursor to learning topology, I would make the case for picking up a little category theory along the way, as it becomes relevant (the "just-in-time category theory" approach). For example, when you learn about the product topology and the topological sum, it's appropriate to devote some effort to understanding the universal properties of the product and the coproduct, and when you start learning about the fundamental group or homology, you'll want to learn what a functor is. It may not be strictly necessary in order to understand the topology, but let me give two reasons why picking up this small amount of category theory may be beneficial, in the particular case of products and coproducts.
1. It motivates the construction of the topological product and the topological sum. Sure, you can construct this topology on the disjoint union of some collection of spaces, making it into the topological sum, but why should you do such a thing? Why is the topological sum a thing worth considering? There are many possible answers, but one of them is that the topological sum is the coproduct in the category of topological spaces and continuous functions. Without even worrying about what that means precisely, it says that the topological sum behaves analogously to coproducts in other categories. So, if you believe that the disjoint union of sets (which is the coproduct in the category of sets) or the direct sum of groups or vector spaces (which are the coproduct in the categories of groups or vector spaces, respectively) are interesting or useful, then probably the topological sum is also interesting or useful.
An even better example of this idea is the product of topological spaces. You form the product of spaces by taking the cartesian product of their underlying sets, and then putting a suitable topology on it. Well, there are (at least) two topologies a reasonable person might try, the product topology and the box topology, which are different if you're taking the product of infinitely many spaces. I don't know about you, but when I learned about these for the first time, I was convinced that the box topology was a more sensible choice (why should we restrict all but finitely many of the open sets to be the entire space??). Well, the product topology is the product in the category of topological spaces, and the box topology isn't.
2. More practically, the universal property is useful for producing continuous maps. Have you ever wanted to define a continuous map from the topological sum to some other space? (If you haven't, you will.) More concretely, let's say you want to build a continuous map
$$
\coprod_{i \in I} X_i \to Z
$$
where the $X_i$'s and $Z$ are some topological spaces. It turns out that the topological sum has exactly the right topology to make this an easy task. All you have to do is produce a continuous map
$$
X_i \to Z
$$
for each $i$! The universal property of the coproduct says that choosing continuous maps $X_i \to Z$ uniquely determines a continuous map $\coprod X_i \to Z$. I'll grant that you can understand and use this universal property without stating it in the language of category theory, but the same universal property applies to coproducts in any category, so phrasing it in categorical language makes clear similarity in the behavior of objects (such as topological sums and direct sums of vector spaces) that otherwise look completely different.
In 3) the phrase ' there exist an open set $U$' should be changed to ' there exist a non-empty open set $U$'. [Otherwise the result is false since we can always take $U$ to be the empty set].
Suppose 3) holds and $E$ is not nowhere dense. Then the interior of $\overline {E}$ is a non-empty open set $A$. [$\overline {E}$ is the closure of $E$]. By 3) there exist a non-empty open set $U$ such that $U \subseteq A$ and $U \cap E =\emptyset$. If $x$ is any point of $U$ then $x \in A \subseteq \overline {E}$. But $U$ is a neighborhood of $x$ which does not intersect $E$. This is a contradiction.
Best Answer
You seem to be confusing things. First $\coprod_{j \in J} X_j$ (the coproduct or topological sum) should be the "disjointified" union $$\bigcup_{j \in J} X_j \times \{j\}$$ and the injections are $\sigma_j(x)=(x,j)$ for all $j \in J$.
Then $\coprod_{j \in J} X_j$ indeed has the final topology $\mathcal{T}_f$ wrt the injections $\sigma_j:(X,\mathcal{T}_j) \to X$ as you describe. Open sets $O$ in $\coprod_{j \in J} X_j$ can be naturally written as $$ O = \bigcup_{j \in J} O \cap \sigma_j[X] = \bigcup_{j \in J} \sigma_j[\sigma_j^{-1}[O]]$$ which we can see as a union of (homeomorphic copies of) open subsets from the $X_j$.
We could also form the union $X = \bigcup_{j \in J} X_j$ and have injections $\iota_j(x)=x$ and get a final topology $\mathcal{T}'_f$ on $X$ wrt those injections. But then we need assumptions on the $X_j$ and their mutual relations to have any nice meaningful topology on $X$. E.g. suppose that $X_1=[0,1]$ in its usual topology and $X_2 = [0,1]$ in the discrete topology. Then, when executing this construction, $X \simeq X_1$. There is no trace of $X_2$ anymore. We cannot distinguish between the $\frac{1}{2}$ that comes from $X_1$ and the one from $X_2$ etc. Open sets in $X$ are not unions of open sets from $X_1$ and $X_2$ any more. This construction is never done in practice; it doesn't seem to be useful. The standard sums (first part) are very common however, mostly as a tool in other results.
One common exception where the product view can be valid: if all $X_j$ are the same topological space, say $\forall j: (X_j, \mathcal{T}_j)=(Y, \mathcal{T}_Y)$. In that case $\coprod_{j \in J} X_j \simeq Y \times J$ where $J$ has the discrete (!) topology and $Y \times J$ has the product topology. The homeomorphism is just the identity, which is well-defined on the set level. Check that it's continuous and open between these topologies..