Question: Suppose $A$ is an open set of $\mathbb{R}^n$ under the ordinary topology and is homeomorphic to $B\subset \mathbb{R}^n$. Is it necessary that $B$ be open too?

Answer: 
Let $X=\mathbb R,$ let $\mathcal S$ be the Lebesgue measurable sets, and let $\mu$ be Lebesgue measure on $(\mathbb R,\mathcal S).$ The function $f(x)=x$ is Borel measurable (since every Borel set is Lebesgue measurable). However there are Lebesgue measurable sets $E$ that are not Borel measurable. For such an $E,$ $f(E)=E$ is not Borel measurable
No. Take a non-measurable subset $S$ of ${\mathbb R}$, equipped with the
$\sigma $-algebra $\mathscr P(S)$, and the counting measure. Then the
inclusion of $S$ in ${\mathbb R}$ is a measurable map, but the image of $S$ under
this map is not a measurable subset of ${\mathbb R}$.