This is a question regarding ordinals and probably requires some background knowledge of ordinal arithmetic. I will list what I hope is the relevant information for this question:
$\omega_1$ is defined to be the first uncountable ordinal. Therefore any ordinal $\alpha < \omega_1$ is countable. The fact $\bigcup \omega_1 = \omega_1$ is supposed to be used to be able to prove the definition of $\omega_1$, so the only things we are allowed to take for granted are: (1) $\omega_1$ is a set; (2) $\omega_1$ is an ordinal; (3) $\omega_1$ is uncountable. So at this stage, we are $\textbf{not}$ allowed to assume that $\omega_1$ is the least uncountable ordinal.
In general (although I do not know if this applies to $\omega_1$), if $A$ is a proper subset of $X$ then the union of set $X$ of sets is $\bigcup A = \{x : x \in A \, \mbox{for} \, A \in X \} = A_1 \cup A_2 \cup \ldots \cup A_n$.
I'm not very sure how to start this proof so any help would be greatly appreciated. Thanks.
It is in fact true that for every limit ordinal, countable or otherwise, that $\bigcup\alpha=\alpha$.
So it would suffice to prove that $\omega_1$ is a limit ordinal. This is true because if $\alpha$ is an infinite ordinal then we can define a bijection between $\alpha$ and $\alpha+1$. So if $\omega_1$ is $\alpha+1$ we have that $|\alpha|=|\omega_1|$ and so $\omega_1$ cannot be the least uncountable ordinals.
Finally, why does it hold $\bigcup\alpha=\alpha$ for limit ordinals? Note that if $\gamma<\alpha$ then $\gamma+1<\alpha$, therefore $\gamma\in\gamma+1$ so $\gamma\in\bigcup\alpha$. So $\alpha\subseteq\bigcup\alpha$. The other direction is even simpler.