Defined the equivalence relation on $\mathbb C^2$ (With the standard Euclidean topology) by $(z_1, z_2)\sim (w_1,w_2)$ iff $z_i=\lambda w_i$ for some $\lambda \in \mathbb C\setminus\{0\}$. Then $\mathbb (C\setminus\{0\} )\setminus\sim$ is homeomorphic to $S^2$.
This kind of result in $\mathbb R^3$ or $\mathbb R^2$ are quite natural to imagine. But in $\mathbb C^2\equiv \mathbb R^4$ is out of my imagination, therefore I got no idea how to define the homeomorphism. Any help will be appreciated.
First, I think your description of the space is typo'd. I think you want to consider $(\mathbb{C}^2 \setminus \{0\})/{\sim}$, which is homeomorphic to $S^2$. With that in mind:
This space is called $\mathbb{C}P^1$.
As you said, $\mathbb{R}P^1 = \mathbb{R}^2 / \sim$, is easier to imagine: the points in $\mathbb{R}P^1$ correspond to possible slopes of lines in $\mathbb{R}^2$. The slope of any line in $\mathbb{R}^2$ is either a real number or $\infty$ (by which we mean that the line is "vertical", i.e. it equals $\mathbb{R} \times \{x\}$ for some $x \in \mathbb{R}$). So you can think of $\mathbb{R}P^1$ as the real line, with a single point at $\infty$ attached. Here are some important exercises:
Once you can do these, $\mathbb{C}P^1$ will not be hard to tackle! The intuition here is that $\mathbb{R}P^1$ is "formed from $\mathbb{R}$ by adding a point at infinity" -- this point at infinity joins the two ends of the number line together to form a circle. In terms of slopes: you can rotate a line with positive slope at constant speed until it has negative slope, passing through verticality in the process.
Now, for $\mathbb{C}P^1$, the idea is the same: the points of $\mathbb{C}P^1$ are possible slopes of lines in $\mathbb{C}^2$. Such a slope is either a complex number or $\infty$. So $\mathbb{C}P^1$ should look like the complex plane, with a point at $\infty$ attached. Maybe now it's clearer that this should look like $S^2$?