Proving this 1-manifold is the Hopf link I encountered this example while reading Scorpan's book "The wild world of 4-manifolds".
Consider the planes
$$\Pi_1 =\{(x,y,0,0) \in \mathbb{R}^4\}$$
$$\Pi_2 =\{(0,0,z,w) \in \mathbb{R}^4\}.$$
I would like to prove that

$H:=\mathbb{S}^3\cap (\Pi_1\cup \Pi_2)$ is the Hopf link.

Of course $K_1 = \mathbb{S}^3\cap \Pi_1$  can be parametrized by
$\{(e^{i\theta},0)\}_{\theta \in \mathbb{R}}\subset \mathbb{C}^2\simeq \mathbb{R}^4$ and similarly $K_2 = \mathbb{S}^3\cap \Pi_2$  can be parametrized by
$\{(0,e^{i\theta})\}_{\theta \in \mathbb{R}}\subset \mathbb{C}^2\simeq \mathbb{R}^4$.
I wanted to show that $K_i$ bounds a disk $D_i$ in $\mathbb{S}^3$ 
and that $D_1$ is pierced by $K_2$ exactly once. I did not manage to find the disks though.
 A: Your notation for $K_1$ is a little weird, because you've got $(e^{i\theta}, 0, 0) \in \Bbb C^2$, which doesn't quite make sense -- it seems to have 3 coordinates. But I think I know what you mean. 
Let's say instead that 
$$
K_1 = \{ (\cos t, \sin t, 0, 0) \in \Bbb R^4 \mid 0 \le t \le 2\pi\},$$ OK? 
Well, the obvious disk bounded by $K_1$ is
$$
D_1 = \{ (r \cos t, r \sin t, 0, 0) \mid 0 \le \theta \le 2 \pi, 0 \le r \le 1 \}.
$$
but unfortunately, that's a disk in $\Bbb R^4$ rather than in $S^3$. 
If you drop down a dimension or two, and look at the unit circle in the plane, and consider its intersection with the $x$-axis, you get two points, $(\pm 1, 0)$, which bound the "disk" which is the horizontal segment between these two points, and not a disk within $S^1$. But you can "lift" this disk to be the upper arc of $S^1$ by replacing $(x, 0)$ with $(x, \sqrt{1-x^2})$, right? Let's try that in $S^3$, and define
$$
E_1 = \{ \left( r \cos t, r \sin t, \sqrt{1-r^2}, 0 \right ) \mid 0 \le t \le 2 \pi, 0 \le r \le 1 \}.
$$
Now that is a disk in the 3-sphere, where you wanted it. 
And if we define $K_2$ in the more or less obvious way, as 
$$
K_2 = \{ (0, 0, \cos s, \sin s) \in \Bbb R^4 \mid 0 \le s \le 2\pi\},$$
then for $K_2$ to intersect $E_1$, we have to have the fourth coordinate be zero, i.e., $s = 0, \pi$, and the third coordinate be positive, which happens only at $s = 0$. So they intersect in the single point $(0,0,1,0)$. 
If you do stereographic projection from $(0,0, -1, 0)$, the the resulting picture looks like the following: 
$K_2$ becomes a straight line -- let's say the vertical line -- and $K_1$ becomes the unit circle in the horizontal plane, and $E_1$ becomes the unit disk in that plane, which intersects $K_2$ at the origin. (Why is $K_2$ a vertical line? Because it passes through the stereographic-projection-center, hence ends up with one point at infinity). 
