If for some $i$, $x_i = x_{i+1}$ then we have nothing to do. Assume now, that $x_i \neq x_{i+1}$, $0 \le i \le n+1$. Consider the polynomial $P(x) = Q_1(x)^2 + \dots + Q_n(x)^2 - 1$. It at most has $2(n-1)$ roots and it is easy to verify that $x_1, \dots, x_n$ are roots. Hence, there is $i$, $1 \le i \le n-1$ such that $P(x) < 0$ for all $x \in (x_i, x_{i+1})$. On the other hand, according to Cauchy-Schwartz inequality,
$$\left( \sum_{k=1}^{n} |Q_k(x)| \right)^2 \le n(Q_1(x)^2 + \dots + Q_n(x)^2) < n.$$
We are done.

Proof Details.

First, note that using Cauchy's inequality, it is sufficient to show that:
$$\left( \frac{(x-x_2)(x-x_3)\dots(x-x_n)}{(x_1-x_2)(x_1-x_3)\dots(x_1-x_n)} \right)^2 + \dots + \left( \frac{(x-x_2)(x-x_3)\dots(x-x_n)}{(x_1-x_2)(x_1-x_3)\dots(x_1-x_n)} \right)^2 \le 1$$
Let's denote this function by $h(x)$ and define $g(x) = h(x) - 1$. This is a polynomial of degree $2n-2$ with positive coefficients, and $x_1, x_2, \dots$ are its roots. If this polynomial is decreasing at $x_n$ and increasing at $x_{i+1}$, then, there are at least two roots in this interval because the function is negative at the beginning of the interval and at the end of the interval, the function is also negative. By contradiction, if it becomes positive somewhere, by Intermediate Value Theorem (IVT) it must have at least two roots in this interval.
If the function is increasing or decreasing at both ends of an interval, it must have at least one root of this interval. For example, if it is increasing, then the sign of function near $x_i$ is positive and near $x_{i+1}$ is negative, and by intermediate value theorem it should have a root. According to the following claim, the function is decreasing at $x_1$ and increasing at $x_n$. Therefore, if we group the points based on whether the function is increasing or decreasing, the number of increasing blocks and decreasing blocks are equal.
The function has one root in any interval where its ends are within a block, and two roots in any interval where its beginning is in a decreasing block and its end is in an increasing block. Therefore, we have at least $n-1$ roots inside intervals together with $x_1, x_2, \dots, x_n$. We get $2n-1$ roots for a polynomial of degree $2n-2$, contradiction.

Lemma.

The function $h$ is decreasing before $x_1$ and increasing after $x_n$.

Proof.

First, note that:
$$\frac{(x-x_2)(x-x_3)\dots(x-x_n)}{(x_1-x_2)(x_1-x_3)\dots(x_1-x_n)}$$
is a polynomial of degree $n-1$ with $n-2$ roots. Its derivative therefore has one root in each interval $(x_i, x_{i+1})$. Since the derivative cannot have any other roots due to its degree, the polynomial itself is monotonic before $x_1$ and after $x_n$. Therefore,
$h$ is also monotone. Since $h$ is an even degree polynomial with a positive leading coefficient, in the first interval, it is decreasing, and in the second interval, it must be increasing. ■