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 $K_{d,n} \leq \sqrt{n} \leq n$. 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.