I'm having trouble understanding the following proof.

If $\displaystyle v_1 , v_2 , ..., v_r$ are vectors in a vector space $\displaystyle V$, then:

(a) The set $\displaystyle W$ of all linear combinations of $\displaystyle v_1 , v_2 , ..., v_r$ is a subspace of $\displaystyle V$.

(b) $\displaystyle W$ is the smallest subspace of $\displaystyle V$ that contains $\displaystyle v_1 , v_2 , ..., v_r$ in the sense that every other subspace of $\displaystyle V$ that contains $\displaystyle v_1 , v_2 , ..., v_r$ must contain $\displaystyle W$.

I understand how to prove (a).

Now I have questions regarding proof (b):

The proof of (b) is given like this:

Each vector $\displaystyle v_i$ is a linear combination of $\displaystyle v_1 , v_2 , ..., v_r$ since we can write

$\displaystyle v_i = 0v_1 + 0v_2+... + 1v_i+...+0v_r$

Therefore, the subspace $\displaystyle W$ contains each of the vectors $\displaystyle v_1 , v_2 , ..., v_r$.

Let $\displaystyle W'$ be any other subspace that contains $\displaystyle v_1 , v_2 , ..., v_r$

Since $\displaystyle W'$ is closed under addition and scalar multiplication, it must contain all linear combinations of $\displaystyle v_1 , v_2 , ..., v_r$. Thus $\displaystyle W'$ contains each vector of $\displaystyle W$

( My Question ) How does the author deduce that $\displaystyle W'$ contains all linear combinations of $\displaystyle v_1 , v_2 , ..., v_r$ in the last sentence of above proof?

How does the statement that "$\displaystyle W'$ is closed under addition and scalar multiplication" make the statement that "$\displaystyle W'$ contains all linear combinations of $\displaystyle v_1 , v_2 , ..., v_r$'' true?