Assume I have an N-dimensional system of first-order autonomous ODEs
$y'(t)=F(y(t))$
with $F:\mathbb{R}^N\rightarrow\mathbb{R}^N$ given as
$F(y)=\begin{pmatrix}f_1(y_1,y_2,\ldots,y_N)\\f_2(y_1,y_2,\ldots,y_N)\\ \vdots\\ f_N(y_1,y_2,\ldots,y_N)\end{pmatrix}.$
I want to prove that the system has a unique solution given some initial condition $Y_0$.
One method is, of course, to employ the theorem of Picard-Lindelöf by showing that $F$ is Lipschitz continuous. Thus I would have to prove that there is some norm $||\cdot ||$ on $\mathbb{R}^N$ and some constant $L>0$ such that
$||F(x)-F(y)||\leq L||x-y||.$
However, depending on the nature of $F$ this may get quite tricky.
Now as far as I understand Picard-Lindelöf can also be used for non-autonomous system. Thus my idea was to choose some class $\mathcal{Z}$ of functions $\mathcal{Z}\subset \{z:\mathbb{R}\rightarrow\mathbb{R}^N\}$ (e.g. $z\in \mathcal{Z}$ is bounded) and to define $N$ non-autonomous ODEs for each $z\in\mathcal{Z}$ as
$y_{i,z}'(t)=\bar{f}_{i,z}(t,y_{i,z}(t))$ with $\bar{f}_{i,z}(t,y)=f_i(z_1(t),\ldots,z_{i-1}(t),y,z_{i+1}(t),\ldots,z_N(t))$
and then show that all of the resulting ODEs have a unique solution $y_z(t)=(y_{1,z}(t),\ldots,y_{N,z}(t))^T$ for all functions in $\mathcal{Z}$ (e.g. by using Picard-Lindelöf). Then I show that $y_z\in\mathcal{Z}$ for all $z\in\mathcal{Z}$, and therefore the original autonomous system must have a unique solution.
My intuition tells me that this proof is flawed. I am first assuming that a solution $y$ to the original solutions must be in $\mathcal{Z}$ and then use the results to prove the assumption (circular reasoning). However, I am not sure what part of the proof is invalidated by this. It seems like existence should still hold, however, uniqueness is not given, as I may have alternative solutions not in $\mathcal{Z}$.
On the other hand, if only uniqueness is invalidated, then wouldn't that mean that, if I could prove that $y\in\mathcal{Z}$ for any solution $y$ to the original system some other way (e.g. by showing that there is a boundary where $y'$ always points inward), then I would get existence and uniqueness? Or is that still circular reasoning in the sense: "IF a solution exists THEN it must be bounded" and "IF i have some bounded $z$ THEN a solution must exist".
EDIT:
Since it was pointed out in the comments, that continously differentiable $f_k$ are sufficient to prove the existence of local solutions, I should clarify two aspects of this:
The $f_k$ I am looking at are not continously differentiable in all arguments. $f_i$ is continously differentiable in the $i$-th argument, but not in the others. In fact the concrete $f_k$ have the structure $f_k(y_1,\ldots,y_N)=g_k(y_1,\ldots,y_N)-y_k|\sum_{i=1,i\neq k}^N w_{k,i} y_k|$ where $g_k$ are linear functions. Since the $g_k$ are linear, I can easily prove Picard-Lindelöf for those, but I fail at the other parts. However, with the approach above I can give an upper bound on the sums, and then the remaining part just also becomes linear.
I also require the existence and uniqueness of global solutions. That is, if I have any other argument to conclude from the existence and uniqueness of local solutions to the existence and uniqueness of global solutions, this would be fine as well. I am thinking of something along the lines of: For any $Y_0$ in a bounded set $\mathcal{Y}$ there exists a local solution on the interval $[0,\epsilon]$. However, because of the repelling boundary the end point of the solution is still in $\mathcal{Y}$ so that I can combine all local solutions into a global one. However, I am not sure this argument works.