Working according to your setup above, you should give $\{T,F\}$ the discrete topology, and $\{T,F\}^I$ the product topology of $|I|$ copies of the discrete topology.
In the product topology on $\{T,F\}^I$, a basis for the open sets is given by
$$\{B_{i_1,\dots,i_k,v_1,\dots,v_k}\mid i_1,\dots,i_k\in I, \ v_1,\dots, v_k\in\{T,F\},\ k \in {\Bbb Z}_{>0}\},
$$
where
$$
B_{i_1,\dots,i_k,v_1,\dots,v_k}=\{s=(s_i)_{i\in I}\in \{T,F\}^I\mid s_{i_1}=v_1,\dots, s_{i_k}=v_k.\}
$$
In other words, the basic open sets are simply those which constrain the values of finitely many coordinates of an element in $\{T,F\}^I$ to be fixed values in $\{T,F\}$.
Suppose that you can determine whether or not an element $s=(s_i)_{i\in I}$ of $\{T,F\}^I$ is a member of a set $Y$ by looking at only finitely many coordinates $s_{i_1}$, $\dots$, $s_{i_k}$ of $s$. Then, $Y$ is open, because it is a union of basic open sets. The complement of $Y$ is also open, for the same reason. Therefore $Y$ is clopen (both closed and open.) Given any wff $f\in\Sigma$, whether an assignment satisfies $f$ or not can be determined by looking at only a finite subset of the basic sentences. So, the set $J_f$ of satisfying assignments to $f$ is clopen.
$\{T,F\}$ is finite, so it's compact. Then, by Tychonoff's Theorem, $\{T,F\}^I$ is compact. If every finite set of wffs is satisfiable, the $J_f$s have the finite intersection property (every finite subset of the $J_f$s has a nonempty intersection.) Each $J_f$ is closed, so by compactness, $\bigcap_{f\in\Sigma} J_f\ne\emptyset$. Therefore, $\Sigma$ is satisfiable. This is the Compactness Theorem.
Specifically - the definition of satisfiability is very similar in modal logic.
This similarity is superficial: there's a crucial way in which modal logic is closer to first-order logic in this respect. In propositional logic a model is basically the same thing as a complete theory: if $\Gamma$ is a maximal complete propositional theory, we just consider the valuation sending a propositional atom $a$ to $\top$ if $a\in\Gamma$ and $\perp$ if $\neg a\in\Gamma$, and check that this is in fact a "model" of $\Gamma$ in the sense of propositional logic.
By contrast, in modal and first-order logic it takes work to go from a complete theory to a model of that theory. Intuitively I'd say that the main point is that semantics in the sense of modal and first-order logic involves things which are not necessarily directly referred to in the language. In modal logic, these are the worlds. A model in the context of modal logic is a family of simple objects (= propositional models) connected to each other in a structured way (the underlying frame). The presence of this additional structure, which isn't explicitly determined by the language, makes the passage from complete theories to models nontrivial and resemble the first-order rather than propositional case.
Best Answer
The intuition is that only a maximal (and finitely satisfiable) set of formulas induces (univocally) the definition of a truth assignment which satisfies such a set.
The maximal extension $\Delta$ of the finitely satisfiable set $\Sigma$ of formulas plays the role of a guidance on how to build the truth assignment that satisfies $\Sigma$. Of course, it is not necessary to appeal for $\Delta$ to build a truth assignment that satisfies $\Sigma$: you can just look at the formulas in $\Sigma$ (possibly they are infinitely many) and try to build a truth assignment that satisfies $\Sigma$, for instance by trials and errors, but can you generalize this method to any finitely satisfiable set $\Sigma$ of formulas? In the proof of the compactness theorem we look for a ''procedure'' that allows us to prove that any finitely satisfiable set $\Sigma$ of formulas is satisfiable by showing a truth assignment that satisfies $\Sigma$, and the ''rules'' of this procedure do not depend on $\Sigma$.
Indeed, a crucial point in the proof of compactness is the property that, given a set $\Delta$ of formulas that is maximal (i.e. for every formula $\alpha$, either $\alpha \in \Delta$ or $\lnot \alpha \in \Delta$) and finitely satisfiable, and given a truth assignment $v$ such that, for any propositional variable $X$: \begin{align} v(X) = \begin{cases} \top &\text{if } X \text{ occurs in } \Delta \\ \\ \bot &\text{otherwise,} \end{cases} \end{align} we have that $\overline{v}(\alpha) = \top$ for every formula $\alpha \in \Delta$ (and hence $\Delta$ is satisfiable).
If $\Delta$ is not maximal, you cannot prove the property above, or at least you cannot prove that by simple induction on the formula $\alpha$. Suppose that $\Delta$ is not maximal and contain the formula $X \lor Y$ but neither $X$ nor $\lnot X$ nor $Y$ nor $\not Y$ are in $\Delta$ ($X$ and $Y$ are distinct propositional variables). So, which is a truth assignment that satisfies $\Delta$?
For each choice, we have to check if it is compatible with the other formulas in $\Delta$. If instead $\Delta$ is maximal, we immediately know which is the good choice (we have just to check whether $X$ and $Y$ are in $\Delta$ or not) and we are guaranteed that such a choice is compatible with the other formulas in $\Delta$ because of the (strong) hypothesis of finite satisfiability.
Summing up, the approach followed by Enderton's textbook to prove compactness theorem is the following (it is not the only one, for instance you can derive compactness from the completeness theorem as well):
We want to prove that every finitely satisfiable set $\Sigma$ of formulas is satisfiable.
We have an easy proof (by induction on the definition of formula) of Point 1 if we add the hypothesis that the set is not only finitely satisfiable but also maximal.
We find a way to extend the finitely satisfiable set $\Sigma$ to a maximal finitely satisfiable set $\Delta$ of formulas, so that we can apply the method of Point 2, which is also a proof of Point 1 since $\Sigma \subseteq \Delta$.