Is the following true?

For any $c\in (0,1)$ there exists $f(c)>0$ such that for any subset $A\subset \{1,2,\dots,n\}$ of cardinality $|A|\geq cn$, the set $$B=\left\{ k \in \{1,2,\dots,n!\} \colon \text{ there is } a \in A \text{ that divides } k\right\}$$ of numbers having at least one divisor in $A$ satisfies $$|B|\geq f(c) n!.$$

Of course, $n!$ may be replaced to the any common multiple of elements of $A$, or we may ask about densities or probabilities.

I do not know the answer even for $A=\{ cn\ $consecutive integers$\}$