I'm trying to prove that every planar graph with no cycles of length $3,4,5$ is $3$-colorable. However, I have no opportunity to receive any validation or correction on it, but it would be very helpful for me to learn about the topic, so any assistance or feedback is appreciated.
Let $G = (V,E)$, $n = |V(G)|$, $e = |E(G)|$, be a planar graph with no cycles of length $3,4,5$.
Claim. $G$ is $3$-colorable.
Proof. We first not that if $G$ has no cycles, then $G$ is a forest and thus $3$-colorable. We may therefore assume $G$ has at leas one cycle. Let $M = \{(e,f) \ | \ e \in E(G), f \in F(G)\}$ where $e$ lies on the boundary of $f$ for every $(e,f) \in M$. Then every edge is on the boundary of exactly two faces, whereas $G$ has at least $6$ edges on the boundary of each face. By double counting thus, $2e = |M| \geq 6f$. By Euler's formula thus
\begin{align*} f = 2 + e - n \iff 2e \geq 6f = 12 + 6e - 6n \iff e \leq \frac{3n}{2} - 3. \end{align*}
We now proceed by induction on $n$. Since $G$ has at least one cycle that cannot be of length $3,4,5$, $n \geq 6$. For $n=6$ we have $G = C_6$, which is $3$-colorable. Assume now $n \geq 7$ and the claim holds for smaller n. Let $v \in V(G)$ such that $d_{G}(v) \leq 2$. Indeed, such vertex exists as otherwise $d_{G}(v) \geq 3 \ \forall v \in V(G)$ and thus by the Handshaking lemma
\begin{align*} e = \frac{1}{2} \sum_{v \in V(G)} d(v) \geq \frac{3n}{2} > \frac{3n}{2} - 3 \end{align*}
which is a contradiction to the result above. Let now $G' = G - \{v\}$ be such coloring. By induction hypothesis, $G'$ is $3$-colorable. Let $c: \ V(G') \mapsto \{1,2,3\}$. We now use $c$ to color $G$ and additionally color $v$ with any color from $\{1,2,3\}$ not already used to color a vertex in $N(v)$. Indeed, this is proper as $d_{G}(v) \leq 2$. Thus $G$ is $3$-colorable and this completes the proof. \qed
Is this correct? Am I doing something wrong?
