STEP paper 1 1999, are my deductions correct?  
I have already proven the first two parts. Here's what I'm not sure about
Assume $x_1 \ne x_2 $
$\implies$ $1+\frac1{x_2} \ne 1+\frac1{x_3}$
$\implies$ $x_2 \ne x_3$
....
$\implies$  $x_n \ne x_1$
$\implies$  $1+\frac1{x_1} \ne x_1$
$\implies$  $(x_1)^2-x_1-1 \ne 0$
Implying it has no real solution. But it has two real solutions. Therefore out assumption is wrong and thus $x_1=x_2$
 A: Your proof only shows that, assuming a solution to the original set of equations with $x_1 \ne x_2$ exists, that solution will also have the property that $(x_1)^2 - x_1 - 1 \ne 0$. 
This is not the same thing as saying that no value of $x$ satisfies $x^2-x-1=0$. It's saying that this equation is incompatible with the system of equations you're trying to solve. In other words, you've shown that if we had $x_1 = \frac{1 +\sqrt5}{2}$ or $x_1 = \frac{1-\sqrt5}{2}$, then we'd get $x_1 = x_2 = \dots = x_n$, so if (iii) does not hold then $x_1$ cannot be either of these values.
But of course there's plenty of other possibilities for $x_1$, so there's still more work to do.

Instead, as one possible intermediate step to get from (ii) to (iii), note that if $$x_1 - x_2 = -\frac{x_2 - x_3}{x_2 x_3}$$ then $$|x_1 - x_2| = \frac1{x_2 x_3} |x_2 - x_3|$$ which tells us that $|x_1 - x_2| \le |x_2 - x_3|$, and in fact that $|x_1 - x_2| < |x_2 - x_3|$ unless $|x_1-x_2|=|x_2-x_3|=0$.
More detail: So either some equality $x_i - x_{i+1}$ holds (in which case you can argue that all such equalities must hold, by an argument similar to what you've given) or we can use the above to say
$$|x_1 - x_2| < |x_2 - x_3| < \dots < |x_n - x_1| < |x_1 - x_2|.$$
But this is a contradiction.
A: So, we have $$x_1-x_2=-\frac{x_2-x_3}{x_2x_3},\ x_2-x_3=-\frac{x_3-x_4}{x_3x_4},\ \cdots\ ,\ x_{n-1}-x_n=-\frac{x_n-x_1}{x_nx_1},\ x_{n}-x_1=-\frac{x_2-x_1}{x_1x_2}.$$
The first two combine to give
$$x_1-x_2=\frac{x_3-x_4}{x_2x_3^2x_4},$$
substituting the third expression in gives
$$x_1-x_2=-\frac{x_4-x_5}{x_2x_3^2x_4^2x_5}.$$
Carrying this on gives finally
$$x_1-x_2=(-1)^{n}\frac{x_1-x_2}{x_1^2x_2^2\cdots x_n^2}.$$
Since all $x_i>1$, $x_1^2\cdots x_n^2\neq 1$, so $x_1-x_2=0$.
