I am having trouble understanding how the inductive hypothesis is being used in a proof of the five colour theorem in Matousek & Nesetril's 'Invitation to Discrete Mathematics'. Since it is the beginning of the proof, I will just quote it quickly (emphasis mine),
We proceed by induction on the number of vertices of the graph $G=(V,E)$. For $|V|\leq5$, the statement holds trivially.
By the results of section 5.3 we know that any planar graph has a vertex $v$ of degree at most 5. If we even have $\deg_G(v)<5$ then consider the graph $G-v$, and apply the inductive hypothesis on it. Assuming that the graph $G-v$ is coloured by colours 1,2,...,5, then we colour the vertex $v$ by some colour $i\in\{1,2,...,5\}$ not occurring among the (at most 4) colours used on the neighbours of $v$. In this way, we get a colouring of $G$ by 5 colours.
I don't understand how the inductive hypothesis is being used. It seems to me that they are saying that by the inductive hypothesis, the graph $G-v$ is 5-colourable. But I don't see how our induction gives us that conclusion.
I am trying to elaborate on my misunderstanding of this proof, but I am finding that I cannot even articulate what it is that I don't understand. So perhaps my real question is could somebody rephrase this part of the proof so as to give me another perspective on what is being said?
The induction is on $|V|$, the number of vertices of the graph $G$ in question. The induction hypothesis is, given any planar graph $G$ with $n$ vertices, then $G$ is $5$-colourable.
As part of the induction step, we assume some planar $G'$ has $n + 1$ vertices. We know we can find a vertex $v$ in $G'$ with $4$ or fewer neighbours, and so we let $G$ be $G'$ with the vertex $v$ (and adjacent edges) removed. Note that $G$ now has $n$ vertices, and is still planar.
By the induction hypothesis, $G = G' - v$ must be colourable with $5$ colours. The rest of the argument follows simply from there. Since $G'$ is a completely arbitrary planar graph on $n + 1$ vertices, we have therefore shown that any planar graph on $n + 1$ vertices is $5$-colourable, which is what we're trying to prove in the induction step.
That's what completes the induction proof. We can do this step multiple times to prove it for planar graphs with $n + 2, n + 3,$ etc. The argument allows us to prove the next case from the current one.