Can anyone provide a proof, plz? I need to understand how the proof works, 10x in advance...
Let $\displaystyle a = \pi-3$ and let $\displaystyle A = \left\{3+\frac{a}{2^n}\;\middle\vert\; n\in\mathbb{N}\right\}$ where $\displaystyle \mathbb{N}=\{0,1,2,\dots\}$. In particular, $\displaystyle \pi\in A$. Then we can construct $\displaystyle f:A\to A$ such that $\displaystyle \pi$ is not in the image of $\displaystyle f$. Namely, define $\displaystyle f\left(3+\frac{a}{2^n}\right)=3+\frac{a}{2^{n+1}}$. Then it is easy to extend $\displaystyle f$ to $\displaystyle [3, 4)$ so that $\displaystyle \pi$ is still not in the image of $\displaystyle f$.
See also Hilbert's paradox of the Grand Hotel.
In general, when we are dealing with infinite sets, the cardinality of many set operations depends only on the cardinality of the larger set. In particular, finite set don't matter compared to countably infinite sets, and those don't matter compared to continuum. So, if B is infinite and $\displaystyle |A|\le |B|$ (the cardinality of A is less than or equal to the cardinality of B), then $\displaystyle |A\cup B| = |B|$ and $\displaystyle |A\times B|=|B|$.