Your interpretations for 1 and 2 are correct (except that $f : A \to \text{codomain}(f)$, not $f : \tau \to \text{codomain}(f)$). But they hide a very important part of the process, and from what you write I cannot tell whether you are aware of this.
A topology $\tau$ must satisfy some axioms: (1) the intersection of any two elements of $\tau$ is an element of $\tau$; (2) the union of any collection of elements of $\tau$ is an element of $\tau$; (3) the entire set, and the empty set, are elements of $\tau$.
Your version of 1, for example, says that $\tau$ is the set of all subsets of $A$ the form $f^{-1}(U)$ where $U$ is an element of the topology of $\text{codomain}(f)$.
This begs a big question: Does $\tau$ satisfy the axioms it is supposed to satisfy?
The answer is: Yes it does, but that requires some proof.
The outline of the proof is: Each of the three axioms (1), (2), and (3) for the topology on $\text{codomain}(f)$ implies the same axiom for $\tau$ itself.
Your interpretation for 3 has an error, the correct statement would be that $B$ is weakly compact if it is compact with respect to the subspace topology on $B$ that is induced from $\tau$.
Added in response to a comment from the OP: I don't know how to express compactness of $B$ directly in terms of $\tau$ other than to repeat the standard definition: for every collection of elements $\{U_i\}_{i \in I}$ of $\tau$ such that $B \subset \cup_i U_i$, there exists a finite subset $\{i_1,...,i_N\} \subset I$ such that $B \subset \cup_{n=1}^N U_{i_n}$. This does not imply that $B \in \tau$. Think about the compact subset $[0,1] \subset \mathbb R$ which is not an open subset of $\mathbb R$ (i.e. is not an element of the usual topology on $\mathbb R$).
Your intuition in your first question is close, but we can be a bit more precise. Remember that a topological space is an ordered pair $(X,\mathcal{T})$, consisting of a set $X$ and a topology $\mathcal{T}$. A metric space on the other hand, is an ordered pair $(X,d)$, consisting of a set $X$ and a metric $d$. So, in that sense, a topological space and a metric space are fundamentally different things. That said, there is a natural way to get from a metric space to a topological space. Since metrics let us define open balls, we can use a metric to define a topology. In the case where the metric $d$ induces the same topology $\mathcal{T}$, we say that the topological space $(X,\mathcal{T})$ is metrizable. This is the case with $\mathbb{R}$ and the usual metric. The topology generated by open intervals, and the topology generated by open balls with the standard Euclidean metric, are the same, and so $\mathbb{R}$ with the open interval topology is metrizable. In this case it doesn't really matter whether we think of having a metric space or a topological space, but depending on what we're doing it may be useful to think of it one way or the other.
You may also be interested in reading about metrizable spaces, and examples of topological spaces which are not metrizable, for example $\mathbb{R}$ with the lower limit topology.
As for your second question, it's important to remember that open and closed are not the only types of sets. We say that a set is closed if its complement is open, but that's not the same as saying that a set which is not open is closed (which is false). For $[0,2)$, the complement is $(-\infty,0)\cup[2,\infty)$, which is not open, and so we know that $[0,2)$ is not closed. Because we also know that $[0,2)$ is not open, we conclude that it is neither open nor closed.
Best Answer
I think the word "family" is somewhat ambiguous. For example, Wikipedia says
Here allowed to contain repeated copies of any given member means what Wikipedia denotes as an indexed family :
The first quotation shows that a family of sets can be understood either as a set of sets or as function $f : I \to P$ where $P$ is a set of sets. This is indeed vague and leaves much scope for interpretation. The same vagueness of notation can be found in many textbooks.
For a given set $X$ we may consider
sets $\tau$ of subsets of $X$, i.e. subsets of $\tau \subset \mathfrak P(X)$ = power set of $X$.
"indexed collections" of subsets of $X$, i.e. functions $\theta : I \to \mathfrak P(X)$.
Each subset $\tau \subset \mathfrak P(X)$ can be canonically identified with the inclusion function $\iota(\tau) : \tau \hookrightarrow \mathfrak P(X)$; this produces a "self-indexed collection" of subsets of $X$. Conversely, each function $\theta : I \to \mathfrak P(X)$ determines the subset $\text{im}(\theta) = \theta(I) \subset \mathfrak P(X)$. Clearly $\text{im}(\iota(\tau)) = \tau$, but in general $\iota(\text{im}(\theta)) \ne \theta$. In fact, for a given $\tau \subset \mathfrak P(X)$ there are many $\theta : I \to \mathfrak P(X)$ such that $\text{im}(\theta) = \tau$. One can even show that the "collection" of all $\theta : I \to \mathfrak P(X)$ such that $\text{im}(\theta) = \tau$ is not even a set, but a proper class.
The standard is to define a topology on a set $X$ as a set of subsets of $X$ satisfying suitable axioms.
If you regard the word family as a synonym for set, then you do not get conflicting definitions. If you regard a family as a function, then a topology on $X$ would be some function $\theta : I \to \mathfrak P(X)$ from an index set $I$ to the power set of $X$ which is often written in the form $\{U_i\}_{i \in I}$ (indexed collection of subsets). You can easily modify the axioms for a topology to obtain similar axioms for an indexed collection of subsets of $X$. It is fairly obvious that $\theta$ satisfies these modified axioms if and ony if $\text{im}(\theta)$ is a topology in the standard sense. The essential disadvantage of the alternative definition is that indices are completely unnessary - many formally distinct families of "indexed topologies" give the same "standard topology".