Proof of $e\leq3v-6$ for planar graphs without Euler's formula

Question

For a short talk for an audience not familiar with graph theory I want to give an informal proof that $e\leq3v-6$ holds in all planar graphs with $v>2$ and I don't want to use Euler's formula. My plan is as follows:

• I'll informally introduce what a (simple, undirected) graph is.
• I'll then introduce planar graphs as "can be drawn without edge crossings".
• Finally, I'll introduce maximal planar as "loses planarity if an edge is added".
• Then I'll justify why maximal planar graphs consist of "triangles" - of cycle graphs $C_3$ that enclose faces of $G$. I think it's OK to do that visually.

Now the main part of my proof. We'll start with a given a drawing of a maximal planar graph $G$ without edge crossings - the "original". We will use the following algorithm to copy this drawing which makes sure that the current (partial) copy is always a subgraph of $G$ consisting of triangles of the original which are enclosed by a closed cycle of edges.

1. Start with one of the inner triangles of $G$.
2. As long as we're not finished, one of the edges of our construction is an outer edge of the current copy but an inner edge of the original. It is thus part of a triangle $T$ we haven't copied yet.
3. If we find such a triangle $T$ where the third vertex is missing, we add this vertex to our copy. We will then also need to add the two edges connecting this vertex with the existing edge (which will now become an inner edge).
4. If we can't find a triangle as described in the last step, we have an outer edge $E$ which needs to be connected to a vertex $V$ which is already a part of the current copy. $V$ must obviously be an outer vertex of the current copy. We can also find $E$ and $V$ in such a way that $V$ is directly connected to one of the endpoints of $E$. (Otherwise, adding the missing edges between $E$ and $V$ would create a cycle graph with more than 3 vertices.) That means we only have to add one edge.

At the start of the algorithm, we have 3 outer vertices (and 3 outer edges). At the end, we must again have 3 outer vertices (and 3 outer edges) as the result is maximal planar. Each step of type 3 adds an outer vertex, each step of type 4 converts an outer vertex to an inner vertex. (Alternatively, each type 3 step increases the number of outer edges by one, each type 4 step decreases it by one.) We will thus need $n$ type 3 steps and the same amount $n$ of type 4 steps. These add $3n$ edges and $n$ vertices to our initial triangle and so we end up with $e=3n+3$ edges and $v=n+3$ vertices, proving $e=3v-6$.

$e\leq3v-6$ for planar graphs is now a simple corollary.

Are there any glaring holes in this proof that I have missed?

Edit: I've added a picture to demonstrate the algorithm. The starting triangle is blue. Triangles added by step 3 are green, triangles added by step 4 are orange.