Proof of Propositional Logic Compactness via truth assignment and satisfiability Could anyone tell me where did I go wrong with my working please?

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.
Among these finite subsets there must be one which contains $P_o$, i.e. $\{P_0\}$. According to $u$, $u(P_0)=T$.
Define a $v_0$ on $P_0$ such that $v_0(P_0)=T$. Therefore $v_0(P_0)=u(P_0)=T.$
(The hint as offered by the textbook does not seem redundant, yet I don't see how setting $v_0(P_0)=T$ won't give us the property we need. This is why I feel so uncomfortable about my working; it's too simple and seems to have misunderstood the question.)

I feel that I am not understanding the question correctly and hence, what I am asked to prove, so let me spell out what I understand.
1) Every finite subset $\Delta$ of $\Gamma$ is satisfiable, but it is unclear whether this extends to $\Gamma$ as it is not made clear whether $\Gamma$ is infinite or not.
2) We have defined a sequence of truth assignments $\{V_n\}_{n\in \Bbb N}  $ where the domain of each $V_n$ is the set $\{P_0, P_1,...P_n\}$.
So if we have $\{V_0\}$ and $\{V_1\}$, their domains would be $\{P_0\}$ and $\{P_0,P_1\}$ respectively.
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 $v_0(P_0)=u(P_0)=T.$



 A: 
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 $\Gamma.$
