Given an interval $k$-coloring of $[0,1]$, define a function $f: S^k \to \Bbb R^k$ as follows ($S^k$ is the $k$-sphere). Let $x = (x_1,x_2,...,x_{k+1})$ be a point on the $k$-sphere $S^k$. Define $z = (z_0,z_1,...z_{k+1})$ as follows :
$$z_0 = 0$$ $$\forall j>0, z_j = \sum_{i=1}^{j}x_i^2$$
For $1 < j < k$, define $f_j(x) = \sum_{i=1}^{k+1} \operatorname{sign}(x_i)m_j(i)$, where $m_j(i)$ is the measure of the $j$th color in the segment $[z_{i-1},z_i]$. Finally, $f(x) = (f_1(x),...,f_k(x))$.
This definition can be found on page $2$ of The Borsuk-Ulam Theorem and bisection of necklaces.
How to prove that $f$ is continuous.
Clearly the $z_j$ are continuous. The $m_j$ are also continuous (even Lipschitz with constant $1$). Hence discontinuity can only occur when some $x_i=0$ from the sign function in the $i$th summand. But in a neighbourhood of such points, $m_j(i)$ is $O(x^2)$,
In effect, $f$ is continuous essentially for the same reason that $x\mapsto |x|\cdot x$ is ...