The user wants me to solve a JEE Mathematics problem about the properties of roots of a quadratic equation.

Step-by-step Solution:

1. Analyze the given quadratic equation and conditions. The equation is $\alpha {x^2} - x + \alpha = 0$. We are given that $\alpha$ is a non-zero real number, and the equation has two distinct real roots, ${x_1}$ and ${x_2}$, which satisfy the inequality $|{x_1} - {x_2}| < 1$.

2. Apply the condition for two distinct real roots. For a quadratic equation $ax^2 + bx + c = 0$ to have two distinct real roots, its discriminant $D = b^2 - 4ac$ must be greater than zero ($D > 0$). For the given equation, $a = \alpha$, $b = -1$, and $c = \alpha$. The discriminant is $D = (-1)^2 - 4(\alpha)(\alpha) = 1 - 4{\alpha^2}$. The condition $D > 0$ implies: $1 - 4{\alpha^2} > 0$ $4{\alpha^2} < 1$ ${\alpha^2} < {1 \over 4}$ This inequality holds for $\alpha$ in the interval $\left( -{1 \over 2},{1 \over 2} \right)$.

3. Apply the inequality involving the roots. The difference between the roots of a quadratic equation is given by the formula $|{x_1} - {x_2}| = {\sqrt{D} \over |a|}$. Substituting the values for our equation: $|{x_1} - {x_2}| = {\sqrt{1 - 4{\alpha^2}} \over |\alpha|}$ We are given the inequality $|{x_1} - {x_2}| < 1$. So, ${{\sqrt{1 - 4{\alpha^2}} } \over {|\alpha|}} < 1$ Since $\alpha \neq 0$ and $D > 0$, both sides of the inequality are positive. We can square both sides without changing the direction of the inequality: ${{1 - 4{\alpha^2}} \over {\alpha^2}} < 1$ Since $\alpha^2 > 0$, we can multiply both sides by $\alpha^2$: $1 - 4{\alpha^2} < {\alpha^2}$ $1 < 5{\alpha^2}$ ${\alpha^2} > {1 \over 5}$ This inequality holds for $\alpha > {1 \over \sqrt{5}}$ or $\alpha < -{1 \over \sqrt{5}}$.

4. Combine all conditions to find the set S. The set $S$ consists of all non-zero real numbers $\alpha$ that satisfy both conditions derived above: Condition 1: $-\frac{1}{2} < \alpha < \frac{1}{2}$ Condition 2: $\alpha < -\frac{1}{\sqrt{5}}$ or $\alpha > \frac{1}{\sqrt{5}}$

   To find the intersection, let's compare the boundary values: $\frac{1}{2} = 0.5$ and $\frac{1}{\sqrt{5}} \approx \frac{1}{2.236} \approx 0.447$. Thus, $\frac{1}{\sqrt{5}} < \frac{1}{2}$ and $-\frac{1}{2} < -\frac{1}{\sqrt{5}}$.

   Finding the intersection of the intervals:

   • For positive $\alpha$: We need $\alpha < \frac{1}{2}$ and $\alpha > \frac{1}{\sqrt{5}}$. This gives the interval $\left(\frac{1}{\sqrt{5}}, \frac{1}{2}\right)$.
   • For negative $\alpha$: We need $\alpha > -\frac{1}{2}$ and $\alpha < -\frac{1}{\sqrt{5}}$. This gives the interval $\left(-\frac{1}{2}, -\frac{1}{\sqrt{5}}\right)$.

   The condition $\alpha \neq 0$ is satisfied by these intervals. Therefore, the set $S$ is the union of these two disjoint intervals: $S = \left( -{1 \over 2}, -{1 \over \sqrt{5}} \right) \cup \left( {1 \over \sqrt{5}}, {1 \over 2} \right)$

5. Check which of the given options are subsets of S.

   • Option A: $\left( - {1 \over 2} - {1 \over {\sqrt 5 }} \right)$. This notation seems to be a typo for the interval $\left( - {1 \over 2}, - {1 \over {\sqrt 5 }} \right)$, as the other options are intervals. Assuming this is the intended interval, $\left( -\frac{1}{2}, -\frac{1}{\sqrt{5}} \right)$ is a component of $S$, and thus a subset of $S$. So, A is correct.
   • Option B: $\left( - {1 \over {\sqrt 5 }},0 \right)$. This interval lies between the two disjoint intervals of $S$. Its intersection with $S$ is empty. So, it is not a subset of $S$. B is incorrect.
   • Option C: $\left( 0,{1 \over {\sqrt 5 }} \right)$. Similar to option B, this interval is not part of $S$. So, it is not a subset of $S$. C is incorrect.
   • Option D: $\left( {{1 \over {\sqrt 5 }},{1 \over 2}} \right)$. This interval is the other component of $S$, and thus a subset of $S$. So, D is correct.

Final conclusion is that options A and D are the correct choices.