When $n \ge 2$, $S_n \ge 2S_{n-1}$ is equivalent to
$$x_n \ge x_1 + \cdots + x_{n-1}. \qquad ①$$
Let $C = \frac{1}{4}x_1$. We will prove
$$x_n \ge C \cdot 2^n,\ n = 1, 2, \dots \qquad \textcircled{2}$$
by induction.

When $n = 1$, it is obviously true. When $n = 2$, we have $x_2 \ge x_1 = C \cdot 2^2$.

When $n \ge 3$, assume $x_k \ge C \cdot 2^k$, $k = 1, 2, \dots, n-1$.
Then from ①, we have
$$\begin{align*} x_n &\ge x_1 + (x_2 + \cdots + x_{n-1}) \\ &\ge x_1 + (C \cdot 2^2 + \cdots + C \cdot 2^{n-1}) \\ &= C(2^2 + 2^2 + 2^3 + \cdots + 2^{n-1}) = C \cdot 2^n. \end{align*}$$
Therefore, ② holds for every $n$.