WLOG suppose that $u_1 + u_2 + \cdots + u_n$ is the subset of $\{u_1, u_2, \cdots, u_m\}$ with their sum of maximal magnitude. WLOG, suppose that $u_1 + u_2 + \cdots + u_n$ is a vector in the $+x$ direction, by simply rotating all the vectors. Let $u_i = \left<x_i, y_i\right>$ for each $1 \le i \le m.$ Observe that for $1 \le i \le n$ we have $x_i \ge 0$ and for $n+1 \le i \le m$ we have $x_i \le 0.$ This is clear from the maximality of $u_1 + u_2 + \cdots + u_n.$ Now, let $\sigma$ be a permutation of $1, 2, \cdots, n$ such that $|y_{\sigma(1)} + y_{\sigma(2)} + \cdots + y_{\sigma(i)}| \le 1$ for each $1 \le i \le n.$ Similarly, let $\rho$ be a permutation of $n+1, n+2, \cdots, m$ such that $|y_{\rho(n+1)} + y_{\rho(n+2)} + \cdots + y_{\rho(i)}| \le 1$ for each $n+1 \le i \le m.$ From here the finish is easy. Simply start by taking $u_{\sigma(1)}$ to be $v_1$, then take $u_{\rho(n+1)}, \cdots$ until the $x-$component becomes negative, and then take $u_{\sigma(2)}$, etc. This guarantees that, at any point, the $x-$component of a partial sum is at most $1$ in magnitude and the $y-$component is at most $2$ in magnitude. We are done. $\square$