By definition $\bigcup_{\omega\ \Omega}\{\omega\}=\{\omega : \exists \omega'\in \Omega\ \text{ with }\ \omega \in \{\omega'\}\}$.
\begin{align*}1 & =\mathbb{P}[\Omega] \tag{axiom 2} \\ & =\mathbb{P}[\bigcup_{\omega\ \Omega}\{\omega\}] \tag{definition of union} \\ &=\sum_{\omega\ \Omega} \mathbb{P}[\{\omega\}] \tag{axiom 3} \\ &=\sum_{\omega\ \Omega} 0 \tag{assumption} \\ &=0.\end{align*}
I did not understand why in the definition $\omega \in \{\omega'\}$ and $\mathbb{P}[\{\omega\}]$ is $0$ by assumption. Is the proof wrong because it is using a wrong definition?
Edit: axioms refer to the three probability axioms
Probability measures are only countably additive, so your "axiom 3" only applies if $\Omega$ is countable.
Your argument is therefore a proof (by contradiction) that if a probability space is countable, then the singleton events can't all have probability 0.