Here is my attempt at an answer - what I propose is to use a more topological argument. As far as I can tell this works provided the flow curves are of class $C^0$, because it essentially only uses the fact they are Jordan curves.

Let $(\gamma_\alpha \mid \alpha \in A)$ be the union of flow lines, indexed by some set $A$. Each one of is a compact, continuous loop $\gamma_\alpha: \mathbf{S}^1 \to \mathbf{R}^2$ with $\gamma(1) = 0$, which moreover bounds an open set denoted $\Omega_\alpha \subset \mathbf{R}^2 \setminus \{ 0 \}$ whose closure includes the origin.

Define an equivalence relation on $A$ where $\alpha_1 \sim \alpha_2$ if $\Omega_{\alpha_1} \subset \Omega_{\alpha_2}$ or $\Omega_{\alpha_2} \subset \Omega_{\alpha_1}$. The properties of flow lines imply that this is an equivalence: indeed for any two $\alpha_1,\alpha_2 \in A$ either $\alpha_1 \sim \alpha_2$ or $\Omega_{\alpha_1} \cap \Omega_{\alpha_2} = \emptyset$. Index the equivalence classes by some second set $B$. For each $\beta \in B$ we can form an open set in $\mathbf{R}^2 \setminus \{ 0 \}$: $U_\beta = \cup_{\alpha \in A, [\alpha] = \beta} \Omega_\alpha$. These sets are disjoint, and their union is $\cup_{\beta \in B} U_\beta = \mathbf{R}^2 \setminus \{ 0 \}$. Now, as $\mathbf{R}^2 \setminus \{ 0 \}$ is connected, this means that $\lvert B \rvert = 1$. In terms of the original set $A$, we conclude that any two $\alpha_1,\alpha_2 \in A$ are equivalent, and the open sets $\Omega_\alpha$ form an increasing union.

Next we wish to exploit the fact that $\mathbf{R}^2 \setminus \{ 0 \}$ is not simply connected, whereas the $\Omega_\alpha$ all are. Let $\Gamma: \mathbf{S}^1 \to \mathbf{R}^2 \setminus \{ 0 \}$ be an arbitrary continuous loop that is not null-homotopic in $\mathbf{R}^2 \setminus \{ 0 \}$. This is covered by the open sets $\Omega_\alpha$, from which we may extract a finite subcover, say $\Gamma \subset \Omega_{\alpha_1} \cup \cdots \cup \Omega_{\alpha_N}$ with $\Omega_{\alpha_i} \subset \Omega_{\alpha_{i+1}}$ for $i = 1,\dots,N-1$. As they are increasing, $\Gamma \subset \Omega_{\alpha_N}$. Because $\Omega_{\alpha_N}$ is simply connected, we can continuously deform the loop $\Gamma$ to a point: this contradicts our initial choice for the loop because the deformation is in $\mathbf{R}^2 \setminus \{ 0 \}$.