There are two parts of your proof that I would consider incorrect, leaving out important details (which I'm sure you were aware of, but didn't write down):
In the $P \implies Q$ direction, Case 2, you only show that $\deg v_0 \ge 2$, which doesn't mean that it's even: $17 \ge 2$ and $17$ is odd. What you really want to argue here is that we use edges out of $v_0=v_k$ once at the very beginning, once at the very end, and twice every time it appears in the middle of the walk (this part is identical to Case 1), for a total of $2n+2$ edges, which is even.
In the $Q \implies P$ direction, when you write
we branch into $w_j$, walk until we reach $u_i$ again and continue the walk
this will not generally produce an Eulerian tour at the end. The closed walk $w_j$ might visit $u_i$ many times, and if you continue along $c$ the first time you return to $u_i$, you might miss out on many edges in $G_j'$. Instead, what you should do in this step is walk the entirety of the closed walk $w_j$, starting at $u_i$ and therefore ending at $u_i$, and then continue along $c$ to $u_{i+1}$.
Aside from that, I just have a couple of comments on clarity.
First, when you begin your induction
Let $P(n)$ be the proposition that if all $v\in V(G)$ have even degree, then a connected graph $G$ has an Euler circuit. We induct on $n=|E(G)|$.
you're not really using the right form for $P(n)$: you haven't made it a statement that depends on $n$. What you really want to define $P(n)$ to be is the proposition that "If $G$ is a connected graph with $n$ edges and all $v \in V(G)$ have even degree, then $G$ has an Euler circuit." (This also avoids the problem that, within the statement, you talk about $V(G)$ before you've said what $G$ is.)
Second, I'm not sure $P(3)$ is the right base case. I'd actually go with $P(0)$ (a graph with one vertex and no edges is connected, all vertices have even degree, and it has a trivial closed walk) because in your induction step, when you delete the cycle $c$, you may well end up with plenty of components that $P(0)$ should apply to. (And if we have $P(0)$, we can use it to prove $P(3)$: remove the cycle of length 3, and you have three components with $0$ edges.)
Third, a minor nitpick: when you're combining all the cycles together, you should only take the closed walk of $G_j'$ the first time you reach a vertex $u_i \in c$ that is a part of $G_j'$. (This is obvious if you know what you're doing, but it needs to be said if you want to describe a correct algorithm.)
But overall, well done! This is a complicated proof to write, with lots of fiddly details (which I didn't fully appreciate myself before I had to teach a recitation on it last semester), and you did a good job hitting all the important points. Don't let the length of my comments make you think otherwise.
Best Answer
You cannot delete edges in the way you want. Consider the graph below:
This graph (the complete bipartite graph $K_{2,3}$) has two vertices of degree $3$ and three vertices of degree $2$. You cannot reduce any of the degree-$3$ vertices to degree $2$, because then one of the degree-$2$ vertices would end up having degree $1$ or $0$.
The standard approach to this problem is to find a cycle greedily: starting at any vertex, take a walk around the graph until you visit some vertex for the second time - revisiting that vertex creates a cycle.
You could also refine the edge-deletion approach you've currently got. But this will be more complicated.
First, whenever you have an edge between two vertices of degree $\ge 3$, delete it. When this ends, every edge has an endpoint of degree $2$. If all vertices have degree $2$, then your argument works. If not, take a vertex $v$ of degree $\ge 3$ and follow edges from $v$ until we reach a vertex of degree $\ge 3$. It is either
Repeat this until we find a cycle or until all our remaining vertices have degree exactly $2$.