Is $\operatorname{span}\{(0,1)\}$ a point in $\mathbb{R}^2$? Is $S_1 = \operatorname{span}\{(0,1)\}$  a point in $\mathbb{R}^2$?
Does $0$ vector belong to $S_1$?

Is $S_2=\operatorname{span} \{(0,1),(0,2)\}$ a line in $\mathbb{R}^2$?
Does $0$ vector belong to $S_2$?

How to determine whether $0$ belongs to the span or not?
The definition of the span of a set $S$, denoted by $\operatorname{span}(S)$, is the set of all finite linear combinations of the elements of $S$. 

In your case of $S_1$, we therefore have that $\operatorname{span}\{(0,1)\} = c \cdot (0,1) = (0,c)$ for scalars $c \in \mathbb{R}$. The scalar can be anything — $c = 1, c = 3.1$ or anything else as long as it is in $\mathbb{R}$. So yes, $0$ is in $S_1$ as we can choose $c = 0$ to get the $0$ vector. So $\operatorname{span}\{(0,1)\}$ is much more than just a single point. If you choose different values for $c$ and plot them you will see that you'll end up with a line going through the origin. As a matter of fact, it will be the $y$-axis.

Now for $S_2$, notice that $\operatorname{span}\{(0,1)\}$ already contains the point $(0,2)$ as if we were to choose $c = 2$ we get the point $2 \cdot (0,1) = (0,2)$. So the point $(0,2)$ in $S_2 = \operatorname{span}\{(0,1), (0,2)\}$ is redundant. However, lets formally prove that $S_1 = S_2$:

By definition, $S_2 = a \cdot(0,1) + b \cdot(0,2)$ for scalars $a,b \in \mathbb{R}$. Note that $$a\cdot(0,1) + b\cdot(0,2) =  (0,a) + (0,2b) = (0,a + 2b) = (a + 2b) \cdot (0,1).$$

So we have again that $S_2$ is the set of all linear combinations of $(0,1)$ and therefore, $S_1 = S_2$.

Can you see that again $0$ is in $S_2$? Can you see also that in general $0$ is always in the span of a set?

Hope this helps!
$S_1$ is a line, connecting the origin to the given point $[0,1]$, and $S_2$ is the same line, because it already contains $[0,2]$, too.

${\rm span\,}(v)=\{\lambda\,v\,\mid\,\lambda\in\Bbb R\}$.

So, $[0,0]$ and $[0,2]$ both are in ${\rm span\,}([0,1])$.