(a) To find the number of real roots of the polynomial $p(x) = x^4 - x^3 - 1$, we will analyze the behavior of the function and use the Intermediate Value Theorem.

1. Evaluate $p(x)$ at specific points:
$$p(-1) = (-1)^4 - (-1)^3 - 1 = 1 + 1 - 1 = 1 > 0$$
$$p(0) = 0^4 - 0^3 - 1 = -1 < 0$$
$$p(2) = 2^4 - 2^3 - 1 = 16 - 8 - 1 = 7 > 0$$

2. Determine the critical points by finding the derivative $p'(x)$:
$$p'(x) = 4x^3 - 3x^2$$
Set $p'(x) = 0$:
$$4x^3 - 3x^2 = x^2(4x - 3) = 0$$
$$x = 0 \quad \text{or} \quad x = \frac{3}{4}$$

3. Analyze the behavior of $p(x)$ around the critical points:
- For $x < 0$, $p'(x) < 0$ (decreasing).
- For $0 < x < \frac{3}{4}$, $p'(x) < 0$ (decreasing).
- For $x > \frac{3}{4}$, $p'(x) > 0$ (increasing).

4. By the Intermediate Value Theorem, since $p(x)$ changes sign between $-1$ and $0$, and between $0$ and $2$, there are exactly two real roots.

(b) To prove the inequalities for the sum $S$ and product $P$ of the real roots, we use the bounds found in the solution.

1. Approximate the roots:
- For the first root $x_1$:
$$p(-0.82) > 0 \quad \text{and} \quad p(-0.81) < 0 \implies -0.82 < x_1 < -0.81$$
- For the second root $x_2$:
$$p(1.38) < 0 \quad \text{and} \quad p(1.39) > 0 \implies 1.38 < x_2 < 1.39$$

2. Sum of the roots $S$:
$$S = x_1 + x_2 > 1.38 - 0.82 = 0.56 > \frac{6}{11}$$

3. Product of the roots $P$:
$$P = x_1 x_2 < (-0.81) \cdot 1.38 = -1.1178 < -\frac{11}{10}$$

Thus, we have shown that $S > \frac{6}{11}$ and $P < -\frac{11}{10}$.

The final answer is $\boxed{ S > \frac{6}{11} }$ and $P < -\frac{11}{10}$.