I was trying to visualize the 3-sphere, $S^3$.
One way is with the help of hopf fibration, as there is a fibre bundle with total space being $S^3$, $$S^1 \hookrightarrow S^3 \rightarrow S^2$$ and this can be thought of as a $S^1$ fiber over $S^2$. The way I visualize a fibre is like drawing the fiber at each point of the base space, so if we have the base space as $S^1$, a circle, and an $S^1$ fiber is thought of as attaching each point of the circle another circle. If this is done in such a way as is usually represented in lectures, what we get is a torus, $T^2$. And also, we know that a 2-torus is nothing but the cartesian product of two circles, i.e. $S^1\times S^1$.
I tend to extend this way of thinking and tend to construct a 3-sphere as the cartesian product of a 2-sphere and a 1-sphere.
My doubt is that whether this way of thinking is right or not. Or is $S^2\times S^1$ is homeomorphic to $S^3$
No, these are not homeomorphic. You can see this by comparing their fundamental groups. The fundamental group of $ S^{3} $ is trivial, so $\pi_1(S^3)=0 $ because $ S^{3} $ is simply connected. Meanwhile $\pi_1(S^2 \times S^1) = \pi_1(S^2) \times \pi_1(S^1) = 0 \times \mathbb{Z} = \mathbb{Z} $