Constant in Komlos conjecture
A simple way of obtaining a lower bound $c\ge\sqrt{2}$ is to consider pairs of vectors $u,v\in\mathbb{R}$. First of all, it makes sense to focus on pairs of unit vectors for which all $\{-1,1\}$-linear combinations are as long as possible (note that this is just an interesting special case, I'm not saying that it is opotimal in any way). This is achieved when $u,v$ are orthogonal, and by checking the possible rotations we find that $u=\frac{1}{\sqrt{2}}(1,1), v=\frac{1}{\sqrt{2}}(1,-1)$ show that $c$ must be at least $\sqrt{2}$.

This example can be generalized to the sets of vectors $V_k=\{2^{-\frac{k}{2}}v_{i,k}\ |\ 0\le i\le k\}\subseteq\mathbb{R}^{2^k}$, where the $j$-th coefficient $(v_{i,k})_j$ of $v_{i,k}$ is $1$ if the $i-th$ binary digit in $j$ is $0$, and $-1$ otherwise.

The $\infty$-norm of any $\{-1,1\}$-linear combination of the vectors in $V_k$ is $(k+1)2^{-\frac{k}{2}}$, which reaches its maximum $\frac{3}{2}$ at $k=2$, with the set of vectors

$V_2=\{\frac{1}{2}(1,1,1,1),\frac{1}{2}(1,1,-1,-1),\frac{1}{2}(1,-1,1,-1)\}$.

There may be better lower bounds, but this is a start.