A decision problem A is NP-hard means that if you can solve A in polynomial in input size, you can solve any NP problem in polynomial time (in input size). The mechanism to convert a problem B to a problem A (in polynomial time) is called a reduction from B to A.
Input: A graph G
Question: Is G 3-colourable (i.e., is $\chi(G)\leq 3)$ ?
The problem 3-COLOURABILITY is NP-hard because there is a polynomial time reduction from 3-SAT to 3-COLOURABILITY and there is a reduction from SAT to 3-SAT. It is proven that if you can solve SAT in polynomial time, you can solve any NP problem in polynomial time (Cook's theorem). Hence, checking if chromatic number is at most 3 is hard and therefore finding chromatic number exactly must be hard as well.
Note: (Read this in case you get interested in reductions)
You can find a reduction from NAE SAT to 3-COLOURABILITY more easily (i think so).