Let $\Delta$ a consistent set of propositional formulas and $PROP$ the set of all possible formulas. Let
$$\Omega = \{\varphi \in PROP: \Delta \vdash \varphi\}$$
Is the consistency of $\Omega$ a consequence of the consistency of $\Delta$? By definition, it is impossible to derive a contradiction from $\Delta$. Furthermore, any $\varphi \in \Omega$ is a consequence of (or a theorem with assumption) $\Delta$. It follows that whatever assignment $f$ makes the formulas in $\Delta$ true must make those in $\Omega$ true as well.
On the other hand, anything is implied by a true formula. In other words, if $\omega$ is assumed to be true, then $\omega \to \gamma$ is true for any $\gamma$. But this implies that $\omega \to \neg \gamma$ is true as well, and hence I find that contradictions can be derived from a given hypothesis. This is what confused me in my reasoning, making me unsure of the reasoning in the preceding paragraph.
$\Omega$ is indeed a consistent set of formulas. In other words, it is indeed possible for all formulas in $\Omega$ to be simultaneously true. This is because every formula $\varphi \in \Omega$ is a proof-theoretic consequence of one or more premsies in a consistent set, that is, $\Delta$. In turn, this implies that every formula $\varphi \in \Omega$ must be true whenever all premises in $\Delta$ are true.
However, it is not the case that anything is implied by a true formula. For instance, $\omega$ may be true, but if $\gamma$ is false, then the conditional formula $\omega \to \gamma$ is false while the conditional formula $\omega \to \neg \gamma$ is true. This stems from the very definition of the logical connective known as the arrow "$\to$." Any formula $P \to Q$ whose main connective is the arrow "$\to$" is false whenever the antecedent $P$ is true and the consequent $Q$ is false; otherwise, the formula is true.