Solution:

The bound on $k$ follows from iterating part (a).
Let $G$ be a graph with chromatic number $n$. Consider a coloring of $G$ using $n$ colors labeled $1,2, \ldots, n$. For $i$ from $1$ to $\left\lceil \log_{2}(n) \right\rceil$, define $G_{i}$ to be the graph on the vertices of $G$ for which two vertices are connected by an edge if and only if the $i$th digit from the right in the binary expansions of their colors do not match. Clearly each of the graphs $G_{i}$ have chromatic number at most $2$, by coloring each node with the $i$th digit of the binary expansion of their color in $G$. Moreover, each edge occurs in some $G_{i}$, since if two vertices match in every digit they are not connected by an edge. Therefore $G_{1} \cup G_{2} \cup \cdots \cup G_{\left\lceil \log_{2}(n) \right\rceil}=G$, and so we have found such a decomposition of $G$.