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.
If $e_1,e_2$ are adjacent then $e_1e_2$ is a path satisfying the conditions. Otherwise, let $u,v$ and $x,y$ be $e_1$'s and $e_2$'s endpoints respectively and consider $G-e_1-e_2$; it has at most three connected components, so at least one $C$ containing two of $u,v,x,y$.
If $C$ contains only $u,v$ or only $x,y$ among the four special vertices, adding $e_1$ and $e_2$ back will not connect $C$ to the rest of the graph, so $G$ is disconnected – contradiction. Hence $C$ contains a path $p$ from one vertex incident to $e_1$ to one vertex incident to $e_2$; $e_1pe_2$ is then a path satisfying the conditions.
Best Answer
There is no guarantee $v_{n+1}=v_1$. For example, take a "bow-tie" graph (i.e., two 3-cycles with a vertex in common) and start at $v_1$ any degree-2 vertex, $v_2$ the cutvertex and now only walk the other triangle for $v_3,v_4,v_5,v_6$.
Instead, note that $v_1,v_2,\dots,v_{n+1}$ are $n+1$ vertices from the set of $n$ vertices of $G$, so by pigeonhole there must be $1\leq i<j\leq n+1$ with $v_i=v_j$. By construction we know $j\neq i+1$ and so ...