I haven't seen that number before!

Retina, 68 61 bytes

+`\b(\w+?)(?<!\b\1 .*)(\w+)$
$1 $2
(?=.* (.+)$(?<=\b\1 .+)) 
<empty>

<empty> is an empty line. Note the trailing space on line 3. You can run the above code from a single file with the -s flag.

Explanation

+`\b(\w+?)(?<!\b\1 .*)(\w+)$
$1 $2

This first step implements rules 1 to 6. It is a regex substitution which is applied repeatedly until the string stops changing (that's what the + is for). In each step we add a single space into the string from left to right (following the rules of the challenge). The regex matches the shortest string of digits which hasn't appeared in the already processed part of the string. We ensure that we're looking at a prefix of the remaining string with the word boundary \b and checking that we can reach the end of the string without passing spaces with (\w+)$. The latter also ensures that we only perform one replacement per step.

(?=.* (.+)$(?<=\b\1 .+)) 
<empty>

This matches any space (which is at the end of the regex), provided that the last segment of the string is the same as any other segment in the string, and replaces them with the empty string. That is, we undo the first step if it resulted in an invalid final segment, implementing rule 7.

Pyth, 24 23 bytes

VzI!}=+kNYaY~k"";?kzjdY

Try it out here.

VzI!}=+kNYaY~k"";?kzjdY    Implicit: z=input(), k='', Y=[], d=' '
Vz              ;          For N in z:
     =+kN                    Append N to k
  I!}    Y                   Is the above not in Y?
          aY k               Append k to Y
            ~k""             After append, reset k to ''
                 ?k        Is k truthy (i.e. not '')
                   z       Print original input
                    jdY    Otherwise print Y joined on spaces

Thanks to @FryAmTheEggman for saving a byte :o)

Pyth, 22 bytes

 faW!}=+kTYY~kdz?tkzsY

Leading space is important.