Constructing sigma algebras in countably many steps

One can find proofs in many set theory texts for the stronger result that, for each countable ordinal level of the Borel hierarchy, there exist Borel sets not belonging to that level. Look in book indexes for "universal set". You can also google Borel + "universal set".

However, I've found very few published proofs that limit themselves to the finite levels situation that most measure theory texts state and which you have stated in your question. For this reason I’ve made note of such proofs when I’ve encountered them, and in case it could be of use to you or others, below are the only three such references that I currently know about.

[1] Patrick Paul Billingsley, Probability and Measure, 3rd edition, Wiley Series in Probability and Mathematical Statistics, John Wiley and Sons, 1995 [reprinted as “Anniversary Edition” in 2012], xiv + 593 pages.

On pp. 30-32 there is a detailed construction of a Borel set that does not belong to any of the finite Borel classes.

[2] Mikls Laczkovich, Conjecture and Proof, Classroom Resource Materials, Mathematical Association of America, 2001, x + 118 pages.

See pp. 98-101 and p. 105 and Exercise 17.7 on p. 106.

[3] Eric M. Vestrup, The Theory of Measures and Integration, Wiley Series in Probability and Statistics, Wiley-Interscience, 2003, xviii + 594 pages.

See Section 1.7 (pp. 28-34). Vestrup’s construction is an expanded version of the construction given in the 3rd edition of Billingsley’s book [1]. For the stronger version involving transfinite Borel classes, see Exercise 2.23 on p. 36 (and Notes on the Problems for 2.23, on p. 555).