We let $G$ be a graph which is $k$-color-critical, meaning $\chi (G) = k$ and removing any vertex results in a graph with a smaller chromatic number.
My attempt has been to suppose that the graph $G$ is disconnected. From here I would then say that $G$ has connected components $G_1,G_2,...,G_n.$ Since I am trying to do a proof by contradiction, I assume that in this case I should be able to remove a single vertex and either end up with a graph that is still $k$-colorable or a graph which has chromatic number $\chi (G-v) < k-1$ for some vertex $v$.
I have shown that if $G$ is $k$-color-critical then by removing a single vertex $v$ from $G$ we get $\chi (G-v)=k-1$.
I am not sure exactly what to do from here.
I am thinking something along the lines of the induced subgraphs $G_1,...,G_n$ and if $G$ is $k$-colorable then the max of the chromatic numbers of the subgraphs would be $k$.