We want to show that for any fixed $n$, $\bigcup_{i=1}^n L_i$ is regular when all $L_i$ are regular.
I understand that this is only true for finite $n$. However, what's wrong with the following proof for infinite $n$ by induction?
Define $L^k$ to be $\bigcup_{i=1}^k L_i$.
For the basis step, we choose $n=1$. Then we have $L^k = L^1 = L_1$ which is regular by definition.
For the inductive hypothesis, let us assume that $L^k$ is regular with $k < n$. Then by the definitions of regular expressions, $L^k \bigcup L_{k+1}$ is also regular. But this is $L^{k+1}$ and so $L^{k+1}$ is regular.
Again, I understand why $n$ cannot be infinite. I am interested as to what point this proof is wrong.
If you want to prove something for all ordinals, you usually need to do transfinite induction and you have two inductive steps to deal with: (i) successor ordinals: from $\phi(\alpha)$ conclude $\phi(\alpha+1)$ and (ii) limit ordinals: from $\phi(\alpha)$ for all $\alpha < \lambda$, conclude $\phi(\lambda)$. If you only do the successor case, you have proved that $\phi(n)$ holds for all finite $n$, but you can't conclude that $\phi(\alpha)$ holds for any infinite $\alpha$ (the canonical counter-example is when $\phi(\alpha)$ is defined to mean $\alpha$ is finite).