First, a counterexample: for $-a_1=a_2=1,g=1,f(x)=1+x^2$, the maximal interval $I_{(0,0)}$ is $[-\pi/4,\pi/4]$, not open!
So I am assuming $f$ is actually defined on $(a_1,a_2)$. The “proof” below isn’t complete, some detail is omitted.
First, it is easy to show that if $F$ is a given antiderivative of $1/f$ (so $F$ monotonous), $G$ some antiderivative of $g$, $x$ is a solution of the differential equation iff $x=F^{-1}(G(t)+c)$ for some constant $c$.
As a consequence, $(t,t_0,x_0) \in D$ iff $t_1 < t,t_0<t_2$ and $a_1<x_0<a_2$ and there exists some $c$ such that $F^{-1}(G(t_0)+c)=x_0$ and $G(t)+c$ is in the domain of $F^{-1}$ (which we denote as $(b_1,b_2)$). Note that in this case, $c=F(x_0)-G(t_0)$.
Thus, $(t,t_0,x_0) \in D$ iff $t,t_0 \in (t_1,t_2)$, $x_0 \in (x_1,x_2)$ and $G(t)-G(t_0)+F(x_0) \in (b_1,b_2)$. This is clearly an “open condition”.
For the second part, just note from the above that $\varphi_{(t_0,x_0)}(t)=F^{-1}(G(t)-G(t_0)+F(x_0))$, hence the global flow is clearly continuous.
You seem to be on the right track, but
- There is no need to restrict $x$ to a line in $\Bbb R^d$.
- $\Vert \dot x \Vert$ is the norm of the derivative, but you seem to be using it for the derivative of the norm. Similarly at $|\dot{\mu}| = |\mu|^2$.
- It is not a-priory clear that $\Vert x(t) \Vert$ is never zero, and therefore differentiable everywhere.
- There is no need to apply Picard-Lindelöf in the neighborhood of the “singularity”.
I would proceed as follows: Assume that a solution $x: [0, \infty) \to \Bbb R^d$ of
$$
\dot x(t) = f(x(t))
$$
with $x(0) \ne 0$ exists. Define
$$
\alpha: [0, \infty) \to \Bbb R, \alpha(t) = \Vert x(t) \Vert^2 = (x(t), x(t)) \, .
$$
Then $\alpha(0) > 0$ and
$$
\dot \alpha(t) = 2 (x(t), \dot x(t)) = 2 (x(t), f(x(t))) \ge 2 \Vert x(t) \Vert^3 = 2 \alpha(t)^{3/2} \, .
$$
In particular, $\alpha$ is monotonically increasing and therefore strictly positive for all $t \ge 0$. It follows that
$$
\dot \alpha(t) > \alpha(t)^{3/2} \, .
$$
Now consider the differential equation
$$
\dot \lambda(t) = \lambda(t)^{3/2} \, , \, \lambda(0) = \alpha(0) > 0
$$
which has the solution
$$
\lambda(t) = \frac{1}{\left( \frac{1}{\sqrt{\alpha(0)}} - \frac t 2\right)^2}
$$
on the interval $0 \le t < \frac{2}{\sqrt{\alpha(0)}} =: T$. Now you can apply the lemma and conclude that
$$
\alpha(t) > \frac{1}{\left( \frac{1}{\sqrt{\alpha(0)}} - \frac t 2\right)^2}
$$
for $0 < t < T$. This is a contradiction, because the right-hand side is unbounded for $t \to T$, whereas the left-hand side has the finite limit $\alpha(T)$.
Best Answer
Since the equation for $x$ does not involve $y$, you can solve it first as a single ODE: $$\dot x = f(x), \quad x(t_0)=x_0\tag1$$ Since $f$ is globally Lipschitz, the Picard theorem implies that the solution of (1) exists for all times, and is unique. Now that $x$ is a known function, the equation for $y$ turns into a linear ODE $$\dot y = g(x)y, \quad y(t_0)=y_0\tag2$$ Here $g(x)$ is a continuous function of $t$, being the composition of two continuous functions. It follows that the solution of (2) exists for all times, and is unique.
Remark #1: "continuous and Lipschitz" is redundant. A Lipschitz function is automatically continuous.
Remark #2, concerning your meta question. The Tumbleweed badge is not completely useless. It serves as a catalog of overlooked questions, and thus draws some (small) amount of attention toward them -- which is better than nothing.