find positive real number x that satisfies $2001=x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor$
By letting $f(x)=x\lfloor x\lfloor x \lfloor x\rfloor\rfloor\rfloor $ we have $$ \lim_{x\to 7^-}f(x) = 7(7(7(7 - 1) - 1) - 1) = 2002 $$ and in a left neighbourhood of $x=7$ our function is a linear function with derivative $(7(7(7 - 1) - 1) - 1)=286.$ Since $f$ is increasing, the problem boils down to solving $$ 286(x-7)+2002 = 2001 $$ which leads to $x=\frac{2001}{286}$.
Using what you have so far, suppose $x= 7 - \epsilon$ where $0\leq \epsilon <1.$ Some computing leads me to believe that $\epsilon$ is quite small. Then
$$x\lfloor x \lfloor x\lfloor x \rfloor \rfloor \rfloor = x\lfloor x \lfloor (7-\epsilon)6 \rfloor \rfloor=x \lfloor x\lfloor 42-6\epsilon\rfloor\rfloor.$$
I might have some case work here, but since I think $\epsilon$ is small, I'll start with the case that $\epsilon < 1/6$ to get the above
$$=x \lfloor x(41)\rfloor = x\lfloor (7-x)41\rfloor = x\lfloor 287 - 41\epsilon \rfloor.$$
Perhaps more case work (I fear there might be 41 cases, but maybe I'll get lucky). Assume $\epsilon < 1/41.$ So the equation becomes
$$2001 = x*286$$
And by golly $x = \frac{2001}{286}$ works.
If you know that $x$ is 'just below' $7$ then, keeping your fingers crossed, just go for it!
The OP's setup allows us to write
$\tag 1 \lfloor x \rfloor = 6$
and you can begin by setting $x$ to $6$ on your 'slider bar'.
Now keep pushing $x \lt 7$ to the right until you can write ($\, 7 \times 6 - 1 = 41\,$)
$\tag 2 \lfloor x \times 6 \rfloor = 41$
Now keep pushing $x \lt 7$ to the right until you can write ($\, 7 \times 41 - 1 = 286\,$)
$\tag 3 \lfloor x \times 41 \rfloor = 286$
You are now left with (the rhs of the OP equation after working from the inside to the outside),
$\tag 4 x \times 286 \lt 7 * 286 = 2002$
Of course if you are saving $x = \frac{286}{41}$ from $\text{(3})$ you can push it further to the right and write
$\tag 5 x \times 286 = 2001$
so that the answer is given by
$\tag 6 x = \frac{2001}{286}$
Extra Credit: Determine if the following two equations have solutions for $x \gt 0$:
$\quad 1996=x⌊x⌊x⌊x⌋⌋⌋$
$\quad 1995=x⌊x⌊x⌊x⌋⌋⌋$
Note that we can generate similar problems.
For example, if we started by saying that $x$ is 'just below' $6$ we can crank out another corresponding 'max integer' $n$ such that
$\quad n = x⌊x⌊x⌊x⌋⌋⌋$
We would find $n$ and then ask the student to solve
$\quad 1037 = x⌊x⌊x⌊x⌋⌋⌋$
The procedure/algorithm being defined can actually be proven to produce well-defined outcomes - there is no reason to plug the 'found' $x$ back into the equation to see that 'it works'.