If $S$ is non-empty subset of the vector space $V$, then $span \{S\}$ is a subspace of the vector space V. Why? 
Proposition Let $S$ be non-empty subset of the vector space $V$. The set $span\{S\}$ is a subspace of the vector space $V$.

I'm assuming that this proposition is correct, but I don't understand why.
To show you what I mean, let's consider an example. 
Let $V$ be vector space, where 
$$V = \{(a,b,0)^{T} \mid a,b \in \mathbb R\}$$

The addition of the vectors is defined as 
$$\mathbf{x} + \mathbf{y} = (a,b,c)^{T} + (e,f,g)^{T} = (a+e, b+f,c+g)^{T}$$
And scalar multiplication as:
$$k \mathbf{x} = k\cdot(a,b,c)^{T} = (ka,kb,kc)^{T}$$

Now consider subspace $S$:
$$S = \{(a,0,0)^{T} \mid a\in \mathbb R\}$$
Suppose we have vectors $\bf v_{1}$ and $\bf v_{2}$ such that 
$$\mathbf {v_{1}} = (1,0,0)^{T}$$
and 
$$\mathbf {v_{2}} = (1,1,1)^{T}$$
Consider linear combination 
$$\tag ! k_{1}\mathbf{v_{1}} + k_{2}\mathbf{v_{2}} = \bf u$$
Where $\bf u$ is arbitrary vector in the subspace $S$. We see that, provided that $k_{2} = 0$, any vector in $S$ can be rewritten as the linear combination $(!)$, or in other words, vectors $\bf v_{1},v_{2}$ span set $S$. However, at the same time, $\bf v_{2}$ is not in $V$, hence proposition fails.

I assume there is some flaw in my reasoning, but where exactly?
 A: $\{v_1,v_2\}$ is not a subset of your vector space $V$, or your subspace $S$, as required by the proposition, and presumably your definition of span (since $v_2\notin V$ and $v_2\notin S$). Thus their span is not a subset and therefore not a subspace.
It doesn't matter that $S, V, v_1$ and $v_2$ are all part of a bigger vector space $\Bbb R^3$.
A: You seem to be greatly confused on a couple of concepts:


*

*Spanning set of a vector space.

*The span of a subset of a vector space.


Let $\mathbf{V}$ be a vector space; and let $S$ be a subset of $\mathbf{V}$. We say that $S$ is a spanning set of $\mathbf{V}$ if every vector in $\mathbf{V}$ can be written as a linear combination of elements of $S$.
Note that to say "spanning set of $\mathbf{V}$", the set must be a subset of $\mathbf{V}$.
On the other hand, let $\mathbf{W}$ be a vector space, and let $T$ be a subset of $\mathbf{W}$. The span of $T$ is the set
$$\mathrm{span}(T) = \{ \mathbf{w}\in\mathbf{W}\mid \mathbf{w}\text{ is a linear combination of elements of }T\}.$$ 
Now, the theorem at hand shows that $\mathrm{span}(T)$ is in fact a subspace of the vector space $\mathbf{W}$. One can show more: $\mathrm{span}(T)$ is the smallest subspace of $\mathbf{W}$ that contains $T$, in the following sense: if $\mathbf{Y}$ is any subspace of $\mathbf{W}$ that contains $T$, then $\mathrm{span}(T)\subseteq \mathbf{Y}$.
With these definitions, a subset $S$ of a vector space $\mathbf{V}$ spans $\mathbf{V}$ if and only if $\mathrm{span}(S) = \mathbf{V}$.
Now, you seem to be confused about whether a spanning set must be a subset of your vector space. Let me dispel that confusion: YES; a spanning set must be a subset of the vector space in question.
Your formulation, where you do not require the set to be contained in your space, is not useful; note that if $\mathbf{V}$ is a vector space, and $S$ is a spanning set for $\mathbf{V}$, then it true that every vector in $\mathbf{V}$ is a linear combination of elements of $S$. But then, if you take a proper subspace $\mathbf{W}$ of $\mathbf{V}$, then of course every vector in $\mathbf{W}$ is also a linear combination of elements of $S$ (say, if you have a set that you can use to get any vector in $\mathbb{R}^3$, then you can also use the set to get every vector on the $xy$-plane, $\mathbf{W}=\{(a,b,0)\mid a,b\in\mathbb{R}\}$. But that is not useful because that means going "outside" of $\mathbf{W}$. 
