Finding cube root of a number less than 1 using binary search
Yes, your instructor's one statement is a flaw.  For 0 < x < 1, the root will lie between x and 1.  This is true for any power in the range (0, 1) (roots > 1).
You can reflect the statement to the negative side, since this is an odd root.  The cube root of -1 <= x <= 0 will be in the range [-1, x].  For x < -1, your range is [x, -1].  It's the mirror-image of the positive cases.  I'm not at all clear why the instructor made that asymmetric partitioning.
I am doing MIT6.00.1x course on edX and the professor tells that "If x<1, search space is 0 to x but cube root is greater than x and less than 1".
There are two cases :

 1. The number x is between 0 and 1

2. The number x is less than 0 (negative)

In both the cases, the cube root of x will lie between x and 1. I understood that. But what about the search space? Will the initial search space lie between 0 and x? I think it is not that. I think the bold text as cited from the lecture is a flaw! Please enlighten me on this.