How can we find the elements?

Question

I want to describe an algorithm with time complexity $O(m)$ that, given a set $M$ with $m$ numbers and a positive integer $p \leq m$, returns the $p$ closest numbers to the median element of the set $M$.

low=m/2-(p-1)
high=(m-1)/2+(p-1)
Algorithm(A,begin,end,low,high){
if (begin<end){
   k=medianofMedians(A,low,high)
   q=hoare(A,start,begin,end,k)
   if (q>low) Algorithm(A,begin,q-1,low,high)
   if (q<high) Algorithm(A,q+1,end,low,high)
}
}

Is it right? Now we have an interval, to which the $p$ closest elements belong, but this interval contains $2p$ elements. How could we find the $p$ closest numbers to the median element of the set $M$?

EDIT:

function medianOfMedians(list, left, right)
 numMedians = ceil((right - left) / 5)
 for i from 0 to numMedians
     // get the median of the five-element subgroup
     subLeft  := left + i*5
     subRight  := subLeft + 4
     if (subRight > right) subRight  := right
     medianIdx  := selectIdx(list, subLeft, subRight, (subRight - subLeft) / 2)
     swap list[left+i] and list[medianIdx]
  return selectIdx(list, left, left + numMedians - 1, numMedians / 2)



hoare(arr,start,end, pivot) 
i,j = start,end 
while i < j: 
while i < j and arr[i] <= pivot: 
i += 1 
while j >= i and arr[j] > pivot: 
j -= 1 
if i < j: 
arr[i],arr[j] = arr[j],arr[i] 
return j  

EDIT 2:

Taking into consideration this algorithm:

Select-k(S, k)
   if n = k then return S
   else i ← k mod 2
   low ←Select(A, 1, n, b n/2c − b k/2c)
   high ←Select(A, 1, n, b n/2c + b k/2c)
   if i = 0 then if low ≤ high
           then high ←Select(A, 1, n, b n/2c+b k/2c−1)
   else low ←Select(A, 1, n, b n/2c − b k/2c + 1)
   for i ← 1 to n do
       if low ≤ S[i] ≤ high then return S[i]

how can we know that we return the right $p$ nearesrt elements to the median element?