Reverse Fibonacci

Python 2, 62 59 bytes

n,l,a,b=input(),[],0,1
while a<=n:l=[a]+l;a,b=b,a+b
print l

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Jelly,  9  8 bytes

ḤḶÆḞṚ>Ðḟ

A monadic link taking a number and returning the list of numbers.

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Note: I am assuming 1s are not both required, if they are the previous 9 byter works: +2ḶµÆḞṚfṖ

How?

ḤḶÆḞṚ>Ðḟ - Link: number, n
Ḥ        - double n (this is to cater for inputs less than 6)
 Ḷ       - lowered range of that -> [0,1,2,...,2*n-1]
  ÆḞ     - nth Fibonacci number for €ach -> [0,1,1,...,fib(2*n-1)]
    Ṛ    - reverse that
      Ðḟ - filter discard if:
     >   -   greater than n?

...the reason this only outputs [1,0] for an input of 1 is that the unfiltered list does not contain fib(2), the second 1.

The 9 byter works by adding two, +2, to form the lowered range [0,1,2,...,n,(n+1)] rather than doubling and then filter keeping, f, any results which are in a popped, Ṗ, version of that same list, [0,1,2,...,n] (the value accessed by making a monadic chain, µ).

Haskell, 52 bytes

p=0:scanl(+)1p
f 1=[1,0]
f n=reverse$takeWhile(<=n)p

Second line is necessary for the n = 1 edge-case. I wasn't able to get around it.