set theory – How many Non-Borel Sets are there?

I’m currently doing a course on measure theoretic probability. In the course I learned that the cardinality of Borel $ sigma $-algebra $ mathscr{B} $ is the same as that of continuum $ mathbb{R} $. But what is the cardinality of the set of Non-Borel Sets, let’s call it $ mathcal{S} $.

My inital guess is along these lines. That since $ mathcal{S} = mathscr{B^{c}} $. We can write that,
$$ mathbb{2^R} = mathcal{S} ~ cup mathscr{B} $$ Assume that $ mathcal{S} $ has the same cardinality as $ mathbb{R} $. Now we know that $ mathscr{B} $ also has the same cardinality as that of $ mathbb{R} $. But their union $ mathbb{2^R} $ has a stricly greater cardinality than $ mathbb{R} $. Again assuming that union of uncountable sets of same cardinality has the same cardinality (I’m aware that this is true for countable sets, but not sure about uncountables), this results in a contradiction.

Hence the set $ mathcal{S} $ must have a strictly bigger cardinality than $ mathbb{R} $. Is this correct?

gn.general topology – A non-Borel union of semi-open units

At the complex level $ mathbb $ Look at the half-open square $$ square = {z in mathbb C: 0 le Re (z) <1, ; 0 le Im (z) <1 }. $$

Watch this for everyone $ z in mathbb $ and $ r in {0,1,2,3 } $ the sentence $ (z + i ^ r cdot square) $ is the moved and rotated square $ square $ with a vertex $ z $,

Problem. Is it true that for every function $ r: mathbb C to {0,1,2,3 } $ there a subset $ Z subset mathbb C $ so that the union of squares
$$ bigcup_ {z in Z} (z + i ^ {r (z)} cdot square) $$is not Borel in $ mathbb $?

Added to edit. As @YCor stated in his comment, the answer to this problem is below $ neg CH $,

An affirmative answer to the problem would result from an affirmative answer to another fascinating one

Problem & # 39 ;. Is it true that for each partition $ mathbb C = A cup B $ either $ A $ contains an innumerable strictly increasing function or $ B $ contains an innumerable severely decreasing function?

Here from a function I understand a subset $ f subset mathbb R times mathbb R = mathbb C $ so for everyone $ x in mathbb R $ the sentence $ f (x) = {y in mathbb R: (x, y) in f } $ contains at most one element.

Added in the next edit. In the discussion with @YCor we came to the conclusion that under CH the answer to both problems is negative. Therefore, both problems are independent of ZFC. Very strange.