Given a planar graph, embedded in the plane, you want to add a number of edges to get a maximal planar graph, and you want to be sure that resulting maximal planar graph won't have vertices with degree $=2$.
From this Wikipedia page:
A simple graph is called maximal planar if it is planar but adding any edge (on the given vertex set) would destroy that property...
The process of adding edges is straightforward - you need to find all faces, bounded by more than $3$ edges (including the external face), and triangulate these faces by adding new edges.
Let's consider the case, which looks problematic to you - a vertex $u$ with degree $2$, for which it's impossible to add a new edge, incident to the $u$. You have two vertices $v$ and $w$, adjacent to the vertex $u$, and two faces, partially bounded by edges $\{v,u\}$ and $\{u,w\}$. If a new edge, incident to the vertex $u$, can't be added, then both these faces should have an edge $\{v,w\}$, which prevents this addition. So, the graph contains two parallel edges $\{v,w\}$, which is not possible, because we consider simple graphs only.
Your argument shows that no eulerian triangulation can be obtained by adding a vertex inside a face in a preixisting triangulation and splitting that face into the 3 triangles with the new vertex.
This does not show that no eulerian triangulations exist, for an example of such a graph consider the octahedron.
The theorem is true and is in fact an if and only if, although you are only asked to prove the easy direction.
If there was a vertex of odd degree you can consider the faces around that vertex, and since it's a triangulation you get a cycle of odd length around that vertex, hence not $3$-colorable.
Best Answer
The proof is not valid. In particular, to add a new vertex, it doesn't just need to join two existing vertices, but ones that are adjacent around the polygon built so far; there is no guarantee that such a vertex exists.
The statement is not always true, either. The smallest counterexample is the Goldner–Harary graph shown below:
In this maximal planar graph, there simply is no cycle of length $n$ that visits every vertex once. A quick way to see that is to look at what happens when we delete the $5$ vertices marked with an $X$ below:
The result is $6$ isolated vertices. However, if you have an $11$-gon and delete any $5$ of its vertices, you'll leave the remaining $6$ vertices in $5$ or fewer pieces; there will be some edges between them left. So this graph cannot have an $11$-gon inside it. (In other words, the Goldner-Harary graph is not Hamiltonian because it's not $1$-tough.)