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Why did the author assume $\overrightarrow{0} \in \mathbb{S}_2$? Because that is false, $\overrightarrow{0}$ does not exist in that set then how is he using it to disprove?

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    The author did not assume that $0\in S$. They showed that $0$ does not satisfy the condition of being in $S$, therefore it is not in $S$, therefore $S$ is not a subspace because every subspace must contain $0$. – 2017-01-12

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There doesn't exist any vector space or its subspace without zero vector. Even the smallest vector space is zero vector itself.

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The author does not assume that $0\in\Bbb{S}_2$.

A subspace of $\Bbb{R}^2$ must contain the zero vector ${\bf 0}\in\Bbb{R}^2$. The author notes that ${\bf 0}\notin\Bbb{S}_2$ (because $0+0\neq1$), so $\Bbb{S}_2$ is not a subspace.

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To prove something to be a subspace, it must satisfy the following 3 conditions:

1) The zero vector must be in $\mathbb{S}_2$. ($\mathbf{0} \in \mathbb{S}_2$)

2) It must be closed under vector addition, (If $\mathbf{u}$ and $\mathbf{v}$ are in $\mathbb{S}_2$, $\mathbf{u}+\mathbf{v}$ must be in $\mathbb{S}_2$)

3) It must be closed under scalar multiplication, (If $\mathbf{u}$ is in $\mathbb{S}_2$ and a scalar $c$ is within $\mathbb{R}^3$, then $c\mathbf{u}$ must be in $\mathbb{S}_2$)

In your case, it does not satisfy the first condition to be a subspace.

If you are uncertain, here is a video about it on Khan Academy, it is easier than it looks!

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    The given space does not hold any of these three conditions. In addition: I was told, that $\mathbb{S}_2 \neq \emptyset$ is (together with condition 3) equivalent. If there is a single element $v$ in the subspace, than $0v=0$ must also be there. – 2017-01-12
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    This is true, I intentionally left out parts **(2)** and **(3)** since condition **(1)** is already not satisfied, so we may already deduce that it is not a subspace of $\mathbb{R}^3$. – 2017-01-12
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    I tell my students to "check easy to check", which in most cases, is exactly the existence of $0$. However, there can be other points, that make this point hard to prove, while he addition-condition can be easy to debunk. – 2017-01-12