Given a bounded subset of $\mathbb{R}$, let $\operatorname{st}(^\ast A)$ be the set of the standard parts of all elements in $^\ast$-transform of $A$, $\overline{A}$ is the closure of $A$.Is it true $\operatorname{st}(^\ast A) = \overline{A}$?
It seems to me it's true $\overline{A} \subseteq \operatorname{st}(^\ast A)$.For every $x \in \overline{A}$, there is a sequence $r = \{r_i\}_{i \in \mathbb{N}} \in ^\ast A$ such that $r_i \in A$ for all $i \in \mathbb{N}$ and $\lim_{k \to \infty} r_k = x$. $\operatorname{st}(r) = x$, because for an arbitary $\epsilon$, $\{ i \in \Bbb{N}:|r_i - x|>\epsilon\}$ is finite.
But what about the other direction?
It appears that you’re using the ultrapower construction. Let $\mathscr{U}$ be the ultrafilter. Suppose that $x\in\operatorname{st}{}^*A$. Then there is $a\in{}^*A$ that is represented by a sequence $\langle a_n:n\in\omega\rangle$ in $A$ with the property that for each $n\in\omega$, $$U_n\triangleq\{k\in\omega:|a_k-x|<2^{-n}\}\in\mathscr{U}\;.$$
For each $n\in\omega$ fix $k(n)\in U_n$; then $\langle a_{k(n)}:n\in\omega\rangle$ is a sequence in $A$ converging to $x$, so $x\in\operatorname{cl}A$.