Can someone explain this proof that $\text{span} \{ \vec u, \vec v \} = \text{span} \{ \vec u, \vec v, \vec w \}$? I came across the following proof as part of studying introductory linear algebra.

Let $$\vec u = \begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}, \vec v  = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \vec w = \begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix}. $$ Prove $\text{span} \{ \vec u, \vec v \} = \text{span} \{ \vec u, \vec v, \vec w \}. $
Let $\vec r \in \text{span} \{ \vec u, \vec v, \vec w \}. $ By definition, $$\vec r = \alpha \vec u + \beta \vec v + \gamma\vec w$$ for some $\alpha, \beta, \gamma\in \mathbb{R}. $ Since $\vec w = \vec u + \vec v$, we see that $$\vec r = \alpha \vec u + \beta \vec v + \gamma(\vec u + \vec v) = (\alpha + \gamma) \vec u + ( \beta + \gamma) \vec v \in \text{span} \{ \vec u, \vec v \}. $$ Hence, $\text{span} \{ \vec u, \vec v, \vec w \} \subseteq \text{span} \{ \vec u, \vec v \}. $
Conversely, let $\vec s \in \text{span} \{ \vec u, \vec v \}, $ so that, by definition, $$ \vec s = a \vec u + b \vec v = a \vec u + b \vec v + 0 \vec w $$ for some $a, b \in \mathbb{R}$. This means $\boldsymbol{ \vec s \in \text{span} \{ \vec u, \vec v, \vec w \} }, $ and therefore $ \boldsymbol { \text{span} \{ \vec u, \vec v \} \subseteq \text{span} \{ \vec u, \vec v, \vec w \} } $.
We conclude that $\text{span} \{ \vec v, \vec u\} = \text{span} \{ \vec u, \vec v, \vec w \} $. 

My problem is with the part of the proof that I've emphasized in bold text. 
The definition of span of a set of a vectors is that it's the set of all possible linear combinations of those vectors. 
But $\vec s = a \vec u + b \vec v + 0 \vec w$ does not represent all specific linear combination of those three vectors, because we are multiplying $\vec w$ by the constant $0$. So how does it follow that $\vec s \in \text{span } \{\vec u, \vec v, \vec w \}$?
(An idea I had while writing this question is that maybe it is because $\vec w = \vec v + \vec u$, and we can always find some linear combination of $\vec v, \vec u$ that gives $0$.)
 A: "But $\vec s = a \vec u + b \vec v + 0 \vec w$ does not represent all specific linear combination of those three vectors"
Correct it does not represent all linear combinations of $\vec u, \vec v, \vec w $. But it is a linear combination, where one of the coefficients happens to be zero, so is in their span. Vectors in $\text{span} \{ \vec u, \vec v \} $ are thus also in $\text{span} \{ \vec u, \vec v, \vec w \} $
A: Read this carefully first:

The definition of span of a set of a vectors is that it's the set of
  all possible linear combinations of those vectors.

Now note that since $w$ is a linear combination of $u$ and $v$ (it's their sum) anything you can write as a linear combination of all three vectors you can write as a linear combination of $u$ and $v$.
That's the only inclusion you have to think about. The text in bold that bothers you says simply that any vector you can write as a linear combination of $u$ and $v$ you can write as a linear combination of all three vectors. Just make the third coefficient $0$.
A: Notice that $$s=a\vec u+b\vec v=(a-k)\vec u
+(b-k)\vec v+k\vec w\in\operatorname{span}(\vec u,\vec v,\vec w)$$
But really whether you chose $k=0$ or $k=1$ or even $k=\frac{1372\ln(\pi)}{\sqrt[3]{\sin(666)}}$ does not matter, all are linear combinations of the three vectors.
In the same way, $\vec w=0\vec u+0\vec v+1\vec w\in\operatorname{span}(\vec u,\vec v,\vec w)$, and you are not bothered with the two coefficients equal to $0$.
Remember that you need to prove $\operatorname{span}(\vec u,\vec v)\subset\operatorname{span}(\vec u,\vec v,\vec w)$, which means for any linear combination of $\vec u,\vec v$ there exists at least one linear combination of $\vec u,\vec v,\vec w$ which is the same vector.
So LHS you have "any" and RHS you have "at least one".
A: It sometimes helps to rewrite a mathematical definition with the correct quantifiers in the correct place:

Given a set of vectors $\{\vec u_1,...,\vec u_n\}$, a vector $\vec s$ is a linear combination of the vectors $\{\vec u_1,...,\vec u_n\}$ if there exist scalars $a_1,...,a_n$ for which the following equation is true: $$\vec s = a_1 \vec u_1 + ... + a_n \vec u_n$$ The span of $\{\vec u_1,...,\vec u_n\}$ is the set all linear combinations of the vectors $\{\vec u_1,...,\vec u_n\}$. 

Thus, your vector $\vec s = a \vec u + b \vec v + 0 \vec w$ is unequivocally in the span of $\{\vec u, \vec v, \vec w\}$.
