See : Dirk van Dalen, Logic and Structure (5th ed - 2013), 2.5 Completeness, page 43 :
Lemma 2.5.11 If $\Gamma$ is [maximally] consistent, then there exists a valuation $v$ such that $v(\psi) = 1$, for all $\psi \in \Gamma$.
In order to show that $\Gamma$ is satisfiable, we define a valuation $v$ on propositional variables such that $v(p_i ) = 1$ if $p_i \in \Gamma$ and $v(p_i ) = 0$ otherwise and then extend it to all formulae.
To show that $\varphi = 1$ iff $\varphi \in \Gamma$, we need induction on complexity of $\varphi$.
For atomic $\varphi$, the claim holds by definition [this is the base case for induction].
If $\varphi$ is $ψ ∧ σ$, then $v(\varphi)=1$ iff $v(ψ) = v(σ) = 1$ iff, by induction hypothesis : $ψ,σ \in \Gamma$, and thus $\varphi \in \Gamma$ (because $\Gamma$ is maximally consistent, and thus it is is closed under derivability). Conversely : $ψ ∧ σ \in \Gamma$ iff $ψ,σ \in \Gamma$ and, again by induction hypothesis : $v(ψ) = v(σ) = 1$ and so $v(\varphi)=1$.
The same for $ψ \lor σ$.
If $\varphi$ is $ψ → σ$, then $v(ψ → σ) = 0$ iff $v(ψ) = 1$ and $v(σ) = 0$ iff, by induction hypothesis : $ψ \in Γ$ and $σ \notin Γ$ iff $ψ → σ \notin Γ$, by the properties of maximal consistency.
This is incorrect: consider the situation where we have two propositional atoms $p$ and $q$, and $\Sigma=\emptyset$ (so doesn't prove anything other than tautologies). According to your definition of $u$, there is no restriction on what truth values can be assigned to each of the following sentences (each of which is undecidable in $\Sigma$):
$p$,
$q$,
$\neg(p\wedge q)$.
However, clearly we can't assign each of them "$\top$." So your approach for defining a valuation for $\Sigma$ needs to be more complicated: it needs to take into account how the various sentences, even those undecidable in $\Sigma$, interact with each other.
Note that this problem would not arise if we assumed additionally that $\Sigma$ was complete: that is, for each $\varphi$, either $\varphi\in \Sigma$ or $\neg\varphi\in\Sigma$. (Indeed, then the third clause of your definition of $u$ would be unnecessary.) So what is immediately true is:
$(*)\quad$ Any complete consistent theory has a model.
So now you need to prove:
$(**)\quad$ Any consistent theory is contained in a complete consistent theory.
It's also worth mentioning for contrast what happens in first-order logic. (In particular, this old question nicely parallels yours.)
In first-order logic, the notion of satisfiability is more complex: a model is not just a truth assignment. We still have "consistency implies satisfiability" (for the right proof system, anyways), but the proof is a bit more complicated. However, it evolves quite nicely from the proof of the completeness theorem for propositional logic; see this exposition by Bezhanishvili.
Best Answer
You only need $\Sigma \vdash \phi \land \lnot \phi$ for one formula $\phi$ to show that $\Sigma$ is inconsistent. Here's how I would answer the problem:
Assume $\Sigma$ is inconsistent. Then $\Sigma$ is certainly not empty. So let $\phi \in \Sigma$ and let $\bot$ be any false formula, e.g., $\phi \land \lnot \phi$, so that $\lnot \phi$ is logically equivalent to $\phi \Rightarrow \bot$. Since $\Sigma$ is inconsistent, $\Sigma \vdash \bot$. By the deduction theorem, $\Sigma - \phi \vdash \phi \Rightarrow \bot$ and hence $\Sigma - \phi \vdash \lnot \phi$.