Let $S$  be the set of all non-zero real numbers  $\alpha$ such that the quadratic equation $\alpha x^2-x+\alpha =0$   has two distinct real roots $x_1$  and $x_2$  satisfy the inequality  $|x_1-x_2|<1$ which of the following interval (s) is /are a subset S?

$$\begin{gathered} Given x_1 and x_2 are roots of \alpha x^2-x+\alpha =0 \\ \therefore x_1+x_2=\frac{1}{\alpha } and x_1{x}_2=1 \\ Also, |x_1-x_2|<1 \\ \Rightarrow {|x_1-x_2|}^2<1\Rightarrow {(x_1-x_2)}^2<1 \\ \left(or\right) {(x_1+x_2)}^2-4x_1{x}_2<1 \\ \Rightarrow \frac{1}{{\alpha }^2}-4<1 \\ \Rightarrow \frac{1}{{\alpha }^2}<5 \\ \Rightarrow 5{\alpha }^2-1>0 \\ \Rightarrow (\sqrt{5}\alpha -1)(\sqrt{5}\alpha +1)>0 \\ \therefore \alpha \in (-\infty ,-\frac{1}{\sqrt{5}})\cup (\frac{1}{\sqrt{5}},\infty ) \mathrm{.....}\left(1\right) \\ Also D>0 \\ \Rightarrow b^2-4ac>0 \\ \Rightarrow 1-4{\alpha }^2>0 \\ \Rightarrow \alpha \in (-\frac{1}{2},\frac{1}{2}) \mathrm{.....}\left(2\right) \\ From Eqns. \left(1\right) \& \left(2\right), we get \\ \alpha \in (\frac{-1}{2},\frac{-1}{\sqrt{5}})\cup (\frac{1}{\sqrt{5}},\frac{1}{2}) \end{gathered}$$