Why is $1 - \frac{1}{1 - \frac{1}{1 - \ldots}}$ not real?

You're attempting to take a limit.

$$x_{n+1} = 1-\frac{1}{x_n}$$

This recurrence actually never converges, from any real starting point. Indeed, $$x_2 = 1-\frac{1}{x_1}; \\ x_3 = 1-\frac{1}{1-1/x_1} = 1-\frac{x_1}{x_1-1} = \frac{1}{1-x_1}; \\ x_4 = x_1$$

So the sequence is periodic with period 3. Therefore it converges if and only if it is constant; but the only way it could be constant is, as you say, if $x_1$ is one of the two complex numbers you found.

Therefore, what you have is actually basically a proof by contradiction that the sequence doesn't converge when you consider it over the reals.

However, you have found exactly the two values for which the iteration does converge; that is their significance.

Alternatively viewed, the map $$z \mapsto 1-\frac{1}{z}$$ is a certain transformation of the complex plane, which has precisely two fixed points. You might find it an interesting exercise to work out what that map does to the complex plane, and examine in particular what it does to points on the real line.

I guess that what you're asking for is how the the imaginary unit, i.e. the square root of $-1$ is involved. Indeed it comes from the known identity between continuous fraction and continuous square roots, i.e. $$ \sqrt{a-b\sqrt{a-b\sqrt{a-b\sqrt{a-b\sqrt{\cdots}}}}}= -\cfrac{a}{b-\cfrac{a}{b-\cfrac{a}{b-\cfrac{a}{\ddots}}}} $$ Then you have in your case $a=1$ and $b=1$ you have $$ \sqrt{1-1\sqrt{1-1\sqrt{1-1\sqrt{1-1\sqrt{\cdots}}}}}= -\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\cfrac{1}{1-\ddots}}}} $$ The solution is the well known solution of the equation $$ x=a/(b+x)$$ which brings to the result you found.

When you substitute $a_n=a_{n+1}=x$ in $$a_{n+1}=1-\frac{1}{a_n}$$ you assume that the sequence converges to a fixed-point.

If this assumption is true (as in the + case), this method will help you find the fixed-point.

However, since the sequence does not converge, the solution of $$x=1-\frac{1}{x}$$ cannot be the fixed-point (since there is none).