Clever proof for showing that if a graph G is critically k-colorable then $\delta (G)\ge k-1$

While reading for my graph theory class, I came across a short - yet curious - proof for the following theorem: if a graph G is critically k−colorable then $\delta (G)\ge k-1$. Here is the proof to the claim:

Suppose (for a contradiction) that G is k-critical and that $v\in V(G)$ satisfies $\text{deg}(G)<k-1$. Then $G-v$ has a $(k-1)-$ coloring, and this coloring extends to a $k-1$- coloring of G. This yields a contradiction.

Everything in this proof makes sense besides one particular item: in the second line, what does the author mean by extending a coloring from a subgraph of G to the whole graph? In addition, why does $G-v$ have a $(k-1)$ coloring that extends to all of G (if that makes any sense)? I understand this may seem like an easy Google search but to be honest I can't find anything helpful and figured someone could provide some insight.

Note that this is not a homework question but simply for going beyond what I am learning in class.

While reading for my graph theory class, I came across a short - yet curious - proof for the following theorem: if a graph G is critically k−colorable then $\delta (G)\ge k-1$. Here is the proof to the claim:

Suppose (for a contradiction) that G is k-critical and that $v\in V(G)$ satisfies $\text{deg}(G)<k-1$. Then $G-v$ has a $(k-1)-$ coloring, and this coloring extends to a $k-1$- coloring of G. This yields a contradiction.

Everything in this proof makes sense besides one particular item: in the second line, what does the author mean by extending a coloring from a subgraph of G to the whole graph? In addition, why does $G-v$ have a $(k-1)$ coloring that extends to all of G (if that makes any sense)? I understand this may seem like an easy Google search but to be honest I can't find anything helpful and figured someone could provide some insight.

Note that this is not a homework question but simply for going beyond what I am learning in class.