DropSort it like it's hot

Haskell, 67 59 58 bytes

(q:r)!x|x<last q=q:r!x|1<2=(q++[x]):r
_!x=[[x]]
foldl(!)[]

Explanation: Given a list of lists (that are already sorted) and a value x, the ! operator will place x at the end of the first list whose last element is less than or equal to x. If no such list exists, the list [x] is placed at the end.

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Husk, 10 bytes

hUmü<¡Ṡ-ü<

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This is a combination of my other Husk answer and xnor's Haskell answer. The duplicate ü< feels clunky, but I don't know how to get rid of it...

Explanation

The function ü< translates to nubBy(>) in Haskell. It traverses a list from left to right, keeping those elements for which no previously kept element is strictly greater. In other words, it performs dropsort. The leftover elements are obtained by taking list difference of the original list and the result of ü<.

hUmü<¡Ṡ-ü<  Implicit input, say x = [2,3,5,4,4,2,7].
     ¡      Iterate
      Ṡ-    list difference between argument
        ü<  and its dropsort: [[2,3,5,4,4,2,7],[4,4,2],[2],[],[],[],...
  m         Map
   ü<       dropsort: [[2,3,5,7],[4,4],[2],[],[],[],...
 U          Prefix of unique elements: [[2,3,5,7],[4,4],[2],[]]
h           Drop last element: [[2,3,5,7],[4,4],[2]]

Haskell, 50 bytes

import Data.List
f[]=[]
f l|r<-nubBy(>)l=r:f(l\\r)

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