As Moishe Kohan pointed out in the comment, $C_1 \cup C_2$ forms the so called Hopf link. The definition via Hopf map is in order.
Recall that the homotopy classes of continuous (pointed) maps $S^3\to S^2$ are in one-to-one correspondence with $\pi_3[S^2]=\mathbb{Z}$. We define the Hopf map (a.k.a. Hopf fibration) as the class represented by $1\in\mathbb{Z}$, up to a prescribed convention on which is $1$ and which is $-1$.
A convenient representative of the Hopf map can be described by using quaternions:
Let $$ z\in S^3\subseteq \mathbb{H} = \{a+bi+cj+dk: a,b,c,d\in\mathbb{R}\} $$
denote a unit quaternion. Then $\overline{z}iz$ is a purely imaginary unit quaternion, which means
$$ \overline{z}iz\in S^2, $$
where we identified $S^2$ with $\{q\in\mathbb{H}: |q|=1,\;\operatorname{Re}q=0\}.$
As a result, we have a mapping $f: S^3\to S^2: z \mapsto \overline{z}iz$ which is well known to represent the unit element $\pm1\in\pi_3[S^2].$
In particular, the set $f^{-1}(pt_1)\cup f^{-1}(pt_2)\subset S^3$ has linking number $1$ whenever $pt_1, pt_2$ are two distinct points in $S^2$; Look up Hopf link.
Now we are in position to show that $C_1\cup C_2$ is such a set. Identifying $\mathbb{R}^4\simeq\mathbb{H}$, we have
$$ C_1=V_1\cap S^3=\{e^{i\theta}\in\mathbb{H}:\theta\in\mathbb{R}\}, \\ C_2=V_2\cap S^3=\{e^{i\theta}j\in\mathbb{H}:\theta\in\mathbb{R}\}.$$
Furthermore, it is an easy exercise to show that
$$ C_1 = f^{-1}(i),\; C_2 = f^{-1}(-i), $$
so that $C_1 \cup C_2$ is a typical example of Hopf link.