Show that not all subsets of $[0, 1]$ are Borel-measurable. Find a description of a non-measurable subset ? Hint: use the result about non-existence of shift-invariant measures on $([0, 1],\mathcal P([0, 1]))$
I struggle to show it because I have shown in a previous exercise (Singletons are included in Borel $\sigma$-algebra on $[0,1]$) that for each $x \in [0, 1]$ we have that $\{x\} \in \mathcal F[0,1]$. Since singletons are Borel-measurable then any subset $S \subseteq [0,1]$ can be written this way $$S=\bigcup_{x \in S} \{x\}$$ So $S$ is Borel-measurable and it contradicts what the exercise says.
If the exercise is not wrong, then how can I show it ?
There are $\mathfrak{c}$ many Borel measurable sets and $2^{\mathfrak{c}}$ many subsets of $[0,1]$. So trivially there are many non-Borel measurable sets.