Is this true?
Let $H$ be singular cohomology. On an arbitrary CW complex $X$, given a filtration $X^0 \subseteq X^1 \subseteq \cdots X^n \subseteq \cdots \subseteq X$. Then $$H^*(X) \cong \varprojlim H^*(X^n)$$
I have looked this up, and it seems to me, this only holds for cohomology theories satisfying the Mittag Leffler condition.
But this has been an essential ingredient in the proof of Leray Hirsch Theorem. On page 12, Part 6.) of this notes.
What am I missing?
This isn't true for arbitrary filtrations. But the filtrations used in the proof of the Leray-Hirsch theorem are very special, such that for fixed $i$, the maps $H^i(X^{n+1})\to H^i(X^n)$ are isomorphisms for all sufficiently large $n$. This in particular implies the Mittag-Leffler condition.
Alternatively (and I think this is the argument the notes you are reading has in mind), these filtrations have the property that for fixed $i$ the map $\pi_i(X_n)\to \pi_i(X)$ is an isomorphism for all sufficiently large $n$. By Hurewicz this then implies the map $H^i(X)\to H^i(X^n)$ is an isomorphism for sufficiently large $n$ which immediately implies the map to the inverse limit is also an isomorphism.