Hypothetical Situation for sorting in $O(n)$ using median finding machine that works in $O(\sqrt{n})$
We don't have to merge two sorted arrays, if we can simply compute the median information and store it in an auxilliary array; by definition, a median of an array $A[l...r]$ is the element in $A[(l+r)/2]$ when $A$ is in sorted order. Given this definition, we can simply use mergesort-style divide and conquer, while keeping track of another array $A'$ with medians of subarrays of $A$; in the end, $A'$ will be the sorted version of $A$:
Procedure Median-Merge-Sort($A,l,r,A'$) {

If $r-l&lt;2$ then manually update $A'[l...r]$ and return; $\leftarrow$ Basecase.
Let $m:=$Median($A,l,r$); $\leftarrow$ Compute a median in $O(\sqrt{n})$-time using the oracle.
Let $A'[(l+r)/2]:=m$;
Median-Merge-Sort($A,l,(l+r)/2,A'$);
Median-Merge-Sort($A,(l+r)/2+1,r,A'$);

}
The overall running time is $T(n)=2T(n/2)+O(\sqrt{n})$, which solves to $O(n)$ using master's theorem. This also tells us we cannot have a median-finder working in $O(\sqrt{n})$ on the comparison model, otherwise we can sort in linear time, which violates the sorting lower bound on the comparison model.