I'd like to show that the class of all limit ordinals, LIM, is cofinal in the class of all ordinals ON.
I thought that I could do this by contradiction and assume that it wasn't cofinal in ON. Then there is an $\alpha \in$ ON such that for all $\beta \in $ LIM: $\beta < \alpha$, i.e. $\beta \in \alpha$ and hence LIM $\subset \alpha$.
This means that LIM is a set so we can consider $\bigcup$ LIM. Then we have $\beta < \bigcup$ LIM for all $\beta \in$ LIM by definition of $\bigcup$ LIM. So LIM $\subseteq \bigcup$ LIM.
$\bigcup$ LIM is either a limit or a successor ordinal.
If it is a limit ordinal then $\bigcup$ LIM $\in$ LIM and by transitivity of ON, $\bigcup$ LIM $\subseteq$ LIM, so $LIM = \bigcup$ LIM in this case which would mean that $\bigcup$ LIM $\in$ LIM $= \bigcup$ LIM which would be a contradiction because sets cannot contain themselves. Hence $\bigcup$ LIM cannot be a limit ordinal.
If $\bigcup$ LIM is a successor ordinal then $\bigcup$ LIM = \beta \cup \{ \beta \} for some $\beta \in $ ON.
Now I'm not sure how to proceed from here. I'm also not sure about whether I'm on the right track with this proof because I think there should be a shorter way to prove this.
Thanks for your help.