This problem and its natural higher-dimensional generalization is connected with the recent MO questions Covering a unit ball with balls half the radius and covering disks with smaller disks : let $K_d$ be the smallest constant such that for any sequence $(z_i)_{i \geq 1}$ of vectors of $\mathbb{R}^d$ of (euclidean) norm at most one, there's some choice of signs $s_i = \pm 1$ such that the partial sums $\sum_{1 \leq i \leq n} s_i z_i$ are all bounded by $K_d$.
Now let $N_d$ be the minimal number of balls of radius $\frac{1}{2}$ needed to cover a ball of radius $1$ (in $\mathbb{R^d}$). I claim that $K_d \leq N_d$.
Proof : Let $K_{d,n}$ be the same constant as $K_d$, but for which we require only the first $n$ partial sums to be bounded by $K_{d,n}$. Then a straightforward averaging argument yields $K_{d,n} \leq \sqrt{n}$. sqrt{n} \leq n$. Now let$n > N_d$. Fixing a covering of the unit ball with$N_d$balls of radius$\frac{1}{2}$, then there must be two distinct$ i < j \leq N_d +1$such that$z_i$and$z_j$lie in the same ball of radius$\frac{1}{2}$, and hence must satisfy$|| z_i - z_j || \leq 1$. If we replace$z_j$by$z_j - z_i$, suppress$z_i$, and then use$K_{d,n-1}$, we get a sequence of signs which achieve$K_{d,n} \leq \max ( N_d, K_{d,n-1} ) $. But Kônig's lemma (for infinite binary trees) gives$K_d \leq \sup_{n} K_{d,n} $, hence the desired result. 1 What follows is not an answer, but is too long for a comment. This problem and its natural higher-dimensional generalization is connected with the recent MO questions Covering a unit ball with balls half the radius and covering disks with smaller disks : let$K_d$be the smallest constant such that for any sequence$(z_i)_{i \geq 1}$of vectors of$\mathbb{R}^d$of (euclidean) norm at most one, there's some choice of signs$s_i = \pm 1$such that the partial sums$\sum_{1 \leq i \leq n} s_i z_i$are all bounded by$K_d$. Now let$N_d$be the minimal number of balls of radius$\frac{1}{2}$needed to cover a ball of radius$1$(in$\mathbb{R^d}$). I claim that$K_d \leq N_d$. Proof : Let$K_{d,n}$be the same constant as$K_d$, but for which we require only the first$n$partial sums to be bounded by$K_{d,n}$. Then a straightforward averaging argument yields$K_{d,n} \leq \sqrt{n}$. Now let$n > N_d$. Fixing a covering of the unit ball with$N_d$balls of radius$\frac{1}{2}$, then there must be two distinct$ i < j \leq N_d +1$such that$z_i$and$z_j$lie in the same ball of radius$\frac{1}{2}$, and hence must satisfy$|| z_i - z_j || \leq 1$. If we replace$z_j$by$z_j - z_i$, suppress$z_i$, and then use$K_{d,n-1}$, we get a sequence of signs which achieve$K_{d,n} \leq \max ( N_d, K_{d,n-1} ) $. But Kônig's lemma (for infinite binary trees) gives$K_d \leq \sup_{n} K_{d,n} \$, hence the desired result.