Let $\Gamma=\Sigma \cup \left\lbrace p_i,i\geq 1 \right\rbrace$ a countable set of propositional formulas. Assume also that for every boolean evaluation $u$ that maps every member of $\Sigma$ to true , there exists an $m$ which depends on $u$ such that $u(p_m)=T$.
It comes as an immediate consequence of compactness lemma that one can expand the set $\Sigma$ with those $p_i$ that are satisfied simultaneously with the members of $\Sigma$ in a disjunction. Meaning : the set $\Sigma \cup\left\lbrace p_1\vee p_2\vee ...\vee p_k \right\rbrace$ is satisfiable.
My question is: Are the members of this disjunction finite?
Saying that there exists an integer $k$ such that $\Sigma\models\left\lbrace p_1\vee p_2\vee ...\vee p_k \right\rbrace$ $(A)$
Or $\Sigma\models \left\lbrace p_1\vee p_2\vee ...\right\rbrace$ $(B)$
Is the statement $(A)$ because of the countability of set $\Gamma$ ?
Suppose that for each $k$ there is a truth assignment for $\Sigma$ which makes all of $p_1,\ldots,p_k$ false. In other words, $\Sigma \cup \{ \lnot p_1, \cdots, \lnot p_k \}$ is satisfiable. This means that every finite subset of $\Sigma \cup \{ \lnot p_i : i \geq 1 \}$ is satisfiable, since for every such finite subset there is a maximal index $k$ such that $p_k$ appears in the subset. By compactness, the infinite set $\Sigma \cup \{ \lnot p_i : i \geq 1 \}$ is satisfiable, contradicting your assumption. We conclude that for some $k$, $\Sigma \vDash p_1 \lor \cdots \lor p_k$.