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.
Since all finite subsets of $\Gamma$ are satisfiable, there must be at least one assignment that makes their constituent propositional variables true. Let $u$ be that assignment.
Yes you seem rather confused here. The issue isn't whether the constituent propositional variables are true, it is whether the statements of $\Gamma$ are true. Satisfiability means that we can choose some truth assignment (where in general some of the propositional variables are true and some are false) that make all the sentences of $\Gamma$ true. For instance if $\Gamma=\{\lnot p_0, \lnot p_0\land p_1\}$ then $\Gamma$ is satisfiable, since the truth assignment $p_0=F, p_1=T$ makes all the statements in $\Gamma$ true. Notice whether $p_0$ or $p_1$ happen to need to be true or false has absolutely no bearing here... all that matters is what the truth values of the statements in $\Gamma$ wind up being in a particular truth assignment, and whether you can find one so that they all come up true. Incidentally, there is no assignment where $p_0$ is true that makes this set satisfiable.
(Of course in this problem we are exploring the case where $\Gamma$ is an infinite set of statements, not a finite one like here, but I wanted a simple concrete example here.)
Also,
3) We are asked to show that if we define a $V_0$ on $P_0$, any finite subset of $\Gamma$ can be satisfied by a $u$ such that $v_0(P_0)=u(P_0)=T.$
That's not what it says. You are asked to show that there is some value $v_0(p_0)$ that may be true or false such that for any finite subset of $\Delta \subset \Gamma$ there is an assignment $u$ with $u(p_0) = v_0(p_0)$ that satisfies $\Delta.$ (Though I'm guessing by some of your other comments that this may have been a typo on your part.) In other words, if for every finite subset $\Delta \subset \Gamma,$ there is a truth assignment satisfying it, then furthermore it is possible to pick all these truth assignments so that they all agree on the truth value of $p_0.$
Edit
The issue here is that while finite satisfiability says for each finite set of statements in $\Gamma$ we can find a truth assignment that satisfies them (actually we can generally find an infinite number), we need to show that there's a single truth assignment that works for all of the statements. Of course for any two finite sets, their union is finite, so there's some truth assignment that works for both, but this doesn't mean we can get to a single truth assignment that works for any infinite set of statements. The trick used in this problem is to step back and focus on finding a partial truth assignment that works for everything, rather than a total assignment. In other words, we demand a finite number of the truth values (say of $p_0,\ldots p_n)$ stay fixed, and show that we can still satisfy all finite subsets $\Gamma$ that way.
We start with just fixing $p_0.$ The logic is that if $p_0=T$ does not work, then that means there is some finite subset $\Delta_0\subset \Gamma$ that can't be satisfied with $p_0=T.$ We claim that $\Gamma$ then must be finitely satisfiable by the restricted set of truth assignments with $p_0=F.$ For if not, then there would be some finite $\Delta_1\subset \Gamma$ that can't be satisfied by an assignment with $p_0=F.$ But then we have a contradiction: $\Delta_0 \cup \Delta_1$ is a finite set of statements that can't be satisfied at all (since they can't be satisfied for any assignment with $p_0=T$ or any assignment with $p_0=F$... and that's all truth assignments). This contradicts the fact that $\Gamma$ is finitely satisfiable.
So we've showed that a value exists that we can fix $p_0$ to without wrecking finite satisfiability. So we can keep it fixed there and do the same argument for $p_1$ to show that there's a value we can fix that at as well without wrecking finite satisfiability. So we can keep $p_0$ and $p_1$ fixed at those values, and run the same argument on $p_2.$ And so on. Iterating this over all the variables $p_i$ we build up a total truth assignment that satisfies every finite subset, and thus all the statements of $\Gamma.$
Best Answer
Let $\{X_i\}_{i\in I}$ be a family of compact spaces, and let $X = \prod_{i\in I} X_i$. Let $C$ be a family of closed subsets of $X$ with the f.i.p.
Notation: For $Y_i\subseteq X_i$, let's write $\tilde{Y_i}$ for the set $Y_i\times \prod_{j\neq i} X_i\subseteq X$. The sets $\tilde{U_i}$ with $U_i$ open in $X_i$ form a subbasis for the topology on $X$, so a basic open set has the form $\tilde{U}_{i_1}\cap \dots \cap \tilde{U}_{i_n}$, where each $U_{i_j}$ is open in $X_{i_j}$. Then a basic closed set has the form $\tilde{V}_{i_1}\cup \dots \cup \tilde{V}_{i_n}$, where each $V_{i_j}$ is closed in $X_{i_j}$.
Let's begin with the standard observation that we can replace $C$ with a family of basic closed sets. Indeed, every closed set is the intersection of all the basic closed sets containing it, so letting $$C' = \{V'\mid V'\text{ basic closed, and }V\subseteq V'\text{ for some }V\in C\}$$ we have $\bigcap_{V\in C} V = \bigcap_{V'\in C'} V'$ and $C'$ also has the f.i.p. So we can assume every closed set in $C$ is a basic closed set.
For every closed set $F\subseteq X_i$, introduce a proposition symbol $P_F^i$. A point of $X$ can be written as $(x_i)_{i\in I}$, and we think of $P_F^i$ as meaning "$x_i\in F$".
Now consider the following propositional theory $T$:
I claim that $T$ is satisfiable. By compactness, it suffices to show that we can find an assignment satisfying all axioms from schemas 2 and 3, together with finitely many axioms from schema 1, corresponding to finitely many $V_1,\dots,V_m\in C$. By the hypothesis that $C$ has f.i.p., we can pick some $x = (x_i)_{i\in I}\in V_1\cap\dots\cap V_m$. For each proposition symbol $P^i_F$, set $P^i_F$ to be true if $x_i\in F$ and false otherwise. It is straightforward to check that this assignment works.
Since $T$ is satisfiable, there is some assignment $t$ from our proposition symbols to $\{\top,\bot\}$ satisfying $T$.
Fix $i\in I$. I claim that $C_i = \{F\subseteq X_i\mid F\text{ closed, and }t(P^i_F) = \top\}$ has non-empty intersection. Since $X_i$ is compact, it suffices to show that $C_i$ has the f.i.p. Let $F_1,\dots,F_n\in C_i$. If $\bigcap_{j=1}^n F_j = \varnothing$, then the axiom $\lnot\bigwedge_{j=1}^n P^i_{F_j}$ is in $T$, so we cannot have $t(P^i_{F_j}) = \top$ for all $1\leq j\leq n$, contradiction.
Thus we can pick $x_i\in \bigcap_{F\in C_i}F$ for all $i\in I$. I claim that $x = (x_i)_{i\in I}\in\bigcap_{V\in C} V$. For each $V\in C$ with $V = \tilde{V}_{i_1}\cup \dots \cup \tilde{V}_{i_n}$, $T$ contains the axiom $P^{i_1}_{V_{i_1}}\lor \dots \lor P^{i_n}_{V_{i_n}}$. So for some $1\leq j\leq n$, $t(P^{i_j}_{V_{i_j}}) = \top$. Then $V_{i_j}\in C_{i_j}$, so $x_{i_j}\in V_{i_j}$, so $x\in V$.
Great! But note that above, we have to use the axiom of choice to pick $x_i\in \bigcap_{F\in C_i}F$ for all $i\in I$. If you want to prove Tychonoff's Theorem from compactness for propositional logic relative to ZF, then you need to additionally assume that each $X_i$ is Hausdorff.
The fix is that when $X_i$ is compact Hausdorff, the set $\bigcap_{F\in C_i}F$ is a singleton $\{x_i\}$, so there is no choice to be made. Suppose $x,y\in \bigcap_{F\in C_i}F$ with $x\neq y$. Since $X_i$ is Hausdorff, we can find disjoint open neighborhoods $U_x$ and $U_y$ of $x$ and $y$, respectively. Let $F_x = X_i\setminus U_x$, and let $F_y = X_i\setminus U_y$. Since $x\notin F_x$, we must have $t(P^i_{F_x}) = \bot$. And since $y\notin F_y$, we must have $t(P^i_{F_y}) = \bot$. But since $U_x\cap U_y = \varnothing$, $F_x\cup F_y = X_i$, so the axiom $P^i_{F_x}\lor P^i_{F_y}$ is in $T$. This is a contradiction.
Note that we only used axiom schema 3 in the choiceless proof for compact Hausdorff spaces. But it didn't hurt to throw these axioms into $T$ in the proof for compact spaces as well.