Prove that $χ(G) ≤ 1 + \text{max}\{\text{deg}_{G} (x): x ∈ V\}$ holds for every (finite) graph $G = (V, E)$.
Let's consider the worst case for graph colouring. To obtain the maximal case, we connect every vertex to every other vertex (the complete graph $K_n$). So, for every $v \in V(G)$ its degree is $n-1$, each has the maximal degree. For each of the $n-1$ neighbours we have to colour them with $n-1$ distinct colours. Additionally, the vertex $v$ must also be coloured using a distinct colour, using altogether $n$ colours.
This gives us the upper bound $χ(G) = 1 + \text{max}\{\text{deg}_{G} (x): x ∈ V\}$, every other possible graph must use fewer edges, so the chromatic number must also be smaller (or equal), because we cannot increase the chromatic number by removing edges.
This proof is not compelling because $K_n$ need not be any kind of worst case. Suppose you have a graph $G$ with $60$ vertices, all of degree $\le 3$. How do you conclude that $G$ is 4-colourable from the fact that $K_{60}$ is 61-colourable? (Or: How do you relate $G$ with its $60$ vertices to $K_4$?)
Please try again (or ask for a hint?)