It has odd cycles, so 2 is not enough; 3 colors suffice with an explicit assignment, matching the bound \(\Delta=3\) of Brooks' theorem (it is neither complete nor an odd cycle).
A proper coloring \(c\) with \(k\) colors of a graph \(G=(V,E)\) assigns to each vertex one of the colors \(1,\dots,k\) so that no two adjacent vertices share a color.
\[c:V\to\{1,\dots,k\},\qquad c(u)\neq c(v)\ \text{ if } \{u,v\}\in E\]