Why is $S^3$ $3$-dimensional? Why is the $3$-sphere $S^3$ $3$-dimensional and not $4$-dimensional? We use the standard definition of $S^{n}=\{(x_0,\ldots,x_n) \in \mathbb{R}^{n+1}\}$.
Note: They had written things like submanifold, but i didnt understand
Another question is $I^2$ homeomorphic to a torus ? 
 A: It is locally $3$-dimensional (it is a $3$-manifold). Globally you would have to define some notion of dimension first. 
No it is not. One reason is the hole of the torus. More formally, these spaces habe different invariants as for example different fundamental groups. 
A: It's locally $3$-dimensional: Each point as a neighborhood homeomorphic (or diffeomorphic, etc.) to $\mathbb{R}^3$. Less formally, each point needs three coordinates to describe it locally: if you know $x_1, x_2, x_3$, then $x_4 = \pm \sqrt{1 - x_1^2 - x_2^2 - x_3^2}$. (It's not uniquely determined by the other $x_i$, which is why the coordinates are only local.) It happens to have a nice embedding in a $4$-dimensional space, but that doesn't make it $4$-dimensional itself. A line in $\mathbb{R}^n$ is still $1$-dimensional, even though it's sitting in an $n$-dimensional space.
Although they're both $2$-manifolds, $I^2$ and the torus $S^1\times S^1$ are not the same. The former is contractible, while $S^1\times S^1$ is not; they have different homology and homotopy; $I^2$ is a manifold with boundary, while $S^1\times S^1$ is closed, and so on.
