It seems like this can be done as follows: suppose WLOG that $\sum_{i=1}^n x_i > 0$. Then, let $S_k = \sum_{i=1}^k x_i - \sum_{i=k+1}^n x_i$; we have $S_n = \sum_{i=1}^n x_i > 1$ and $S_0 = -\sum_{i=1}^n x_i < -1$. Suppose that we never have $-1 \le S_k \le 1$. Then let $k_0$ be the maximal value of $k$ for which $S_{k_0} < 0$; we know $k_0$ exists as $k = 0$ certainly satisfies the given property; additionally, $k_0 \ne n$ as $k = n$ does NOT satisfy the property, so $S_{k_0 + 1}$ is defined. By the assumption, we have $S_{k_0} < -1$, and as $k_0$ was the maximal value, we must have $S_{k_0 + 1} \ge 0$ and thus by the assumption $S_{k_0 + 1} > 1$. Thus $2 < S_{k_0 + 1} - S_{k_0} =$ $\left(\sum_{i=1}^{k_0+1} x_i - \sum_{i=k_0+2}^n x_i\right) - \left(\sum_{i=1}^{k_0} x_i - \sum_{i=k_0 + 1}^n x_i\right) =$ $2x_{k_0+1} \le 2|x_{k_0+1}| \le 2$ which is sort of false. Thus there exists a $k$ with the desired property, i.e. that $|S_k| \le 1$.

wow beautiful problem and solution!

Define $$S_k = \left|\sum_{i=1}^k x_i - \sum_{i=k+1}^n x_i\right|.$$In particular, denote $S_0 = -\sum x_i$ and $S_n = \sum x_i$. WLOG suppose that $S_n > 1$; then, we necessarily have $S_0 < -1$. Now, assume for the sake of contradiction that $|S_k| > 1$ for all $k$. Then there must exist some $k, k+1$ such that $S_k < -1$ and $S_{k+1} > 1$. But then $$2|x_{k+1}| > 1-(-1) = 2 \iff |x_{k+1}| > 1,$$which is a contradiction.