Reduction from Maximum Independent Set to Independent Set Decision Problem
It is certainly possible to do so, as every $\mathcal{NP}-$complete problem is self-reducible.

Let us first remember the definition of decisional maximum independent set: given an undirected graph $G=(V,E)$ and $k \in \mathbb{N}$, does there exist $S \subseteq V$ with $|S| = k$ such that $\forall x,  y \in S \ . \ (x, y) \notin E$ ?

By a standard binary search argument, we can find the size of the maximum independent set with a polynomial number of queries to that oracle.

Let $s$ be the size of the maximum independent set of $G$. We pick some $v \in V$ and we use our construction to determine the size of the maximum independent set of $G \setminus v$. That can be either $s$ or $s-1$. If it's $s$, then there exists a maximum independent set in $G$ that does not include $v$, therefore we can exclude $v$ from our search. 

Eventually, we'll reach a point where we can't exclude any other node. That means that the remaining nodes form a maximum independent set.