Let's say we have 3 vectors
$$S=\{(1,2),(3,1),(6,5)\}$$
It has 3 elements hence it belongs to $\mathbb{R}^3$
Is the span of these elements a Subspace of
i) $\mathbb{R}^4$? ii) $\mathbb{R}^3$? iii) $\mathbb{R}^2$?
I am really confused right now is it a subspace of the above or is it not? Or is it just that the example that I took is wrong?
$S$ is a set containing three vectors, each of these belonging to $\mathbb{R}^2$ (not $\mathbb{R}^3$), because they have two real components. Moreover, every two vectors in $S$ are linearly independent (for instance, $(1,2)$ and $(3,1)$ are linearly independent), so $S$ spans $\mathbb{R}^2$.
However, $S$ is not a basis of $\mathbb{R}^2$, since a basis is a set where all vectors are linearly independent, which is clearly not the case. You may verify this by checking that there exist constants $a$, $b$ and $c$ not all zero that satisfy
$a(1,2)+b(3,1)+c(6,5)=(0,0)$
Furthermore, it is obvious that $S$ may not be a basis of $\mathbb{R}^2$, because it contains more than $2$ vectors and a basis of $\mathbb{R}^n$ must contain exactly $n$ vectors (try to prove this!).
In this context, it makes no sense to speak about $\mathbb{R}^3$ or $\mathbb{R}^4$, because the vectors we are working with are defined in $\mathbb{R}^2$. Anyway, you could use the vectors in $S$ to construct a subspace of $\mathbb{R}^3$ (or $\mathbb{R}^4$), by adding one (or two) extra component(s) to each of them, but the subspace that they generate (and its dimension) would depend on your choice for the new component(s) (which is completely arbitrary). For instance,
$V=\{(1,2,0),(3,1,0),(6,5,0)\}$
spans a subspace of $\mathbb{R}^3$ with dimension $2$.
However,
$U=\{(1,2,1),(3,1,2),(6,5,3)\}$
spans $\mathbb{R}^3$ (it is a basis of $\mathbb{R}^3$).