If A is not $F_\sigma$ set, then subset of A is not $F_\sigma$ set

Question

How to prove, that statement "If A is not $F_\sigma$ set, then subset of A is not $F_\sigma$ set." is true or false?

It seems to me to be false. Because by taking contrapositive statement you get "If subset of A is $F_\sigma$ set, then A is $F_\sigma$ set." Let $A={([0,1]\cap\mathbb Q)\cup([2,3]\cap\Bbb I)}$ and $B$ be a subset of $A$ such that $B={([0,1]\cap\mathbb Q)}$. Then implication doesn't hold, because $B$ is $F_\sigma$ set, while $A$ isn't. Is it correct?

Thanks for help.

Answer 1

It's false: the irrationals are not an $F_\sigma$ but contains $\mathbb{Z}\sqrt{2}$ which is closed and countable (so doubly an $F_\sigma$)...

You need not pick such fancy examples, simpler ones will do.