Denote $P(x) = 2x^n - c_{n-1}x^{n-1} + c_{n-2}x^{n-2} - \dots - c_1x + 2$. Since $c_i \in (0, 2)$ for all $1 \le i \le n-1$, it follows $P(x) \ge 2 > 0$ for all $x \le 0$. It remains to prove $P(x) > 0$ also for all $x > 0$. Let $c_i = 1 + \epsilon_i$, so $\sum_{i=1}^{n-1} |\epsilon_i| < 1$. Then for $x > 0$ we have
$$P(x) = (x^n - x^{n-1} + x^{n-2} - \dots - x + 1) + x^n + 1 + \sum_{i=1}^{n-1} (-1)^i \epsilon_i x^i.$$
We have $x^n - x^{n-1} + x^{n-2} - \dots - x + 1 = \frac{x^{n+1}+1}{x+1} > \frac{1}{2} > 0$ since $2x^{n+1} - x + 1 > 0$ (for $x \ge 1$ we have $x^{n+1} - x \ge 0$ while for $x \le 1$ we have $1 - x \ge 0$).

For $0 < x \le 1$ we have (by the triangle inequality)
$$\left| \sum_{i=1}^{n-1} (-1)^i \epsilon_i x^i \right| \le \sum_{i=1}^{n-1} |(-1)^i \epsilon_i x^i| \le \sum_{i=1}^{n-1} |\epsilon_i| < 1,$$
and so
$$x^n + 1 + \sum_{i=1}^{n-1} (-1)^i \epsilon_i x^i \ge x^n + 1 - \left| \sum_{i=1}^{n-1} (-1)^i \epsilon_i x^i \right| > x^n > 0.$$
For $1 \le x$ we have essentially the same argument, but with the role of $x$ above replaced by $1/x$:
$$\left| \sum_{i=1}^{n-1} (-1)^i \epsilon_i (1/x)^{n-i} \right| \le \sum_{i=1}^{n-1} |(-1)^i \epsilon_i (1/x)^{n-i}| \le \sum_{i=1}^{n-1} |\epsilon_i| < 1,$$
and so
$$x^n+1+\sum_{i=1}^{n-1}(-1)^i \epsilon_i x^i = x^n \left( 1 + (1/x)^n + \sum_{i=1}^{n-1} (-1)^i \epsilon_i (1/x)^{n-i} \right) \ge x^n \left( 1 + (1/x)^n - \left| \sum_{i=1}^{n-1} (-1)^i \epsilon_i (1/x)^{n-i} \right| \right) > 1 > 0.$$
Thus $P(x) > \frac{1}{2} > 0$, with room to spare, so we may either relax the condition $\sum_{i=1}^{n-1} |c_i - 1| < 1$ or lower the free term of $P(x)$ to $\frac{3}{2}$.