Prove that $\operatorname{span}(v_1,v_2, \ldots, v_n)$ is subspace Prove that for any vectors $v_1,v_1,\dots,v_n$ from space $V$ above body $K$, $\operatorname{span}(v_1,v_2, \ldots, v_n)$ is subspace of $V$. 
After I read this again, I think I should just prove that that it is closed under addidtion and multiplication. Right? What laws should I use? I give it a try:
$ c ( \alpha v_1+\cdots+\alpha v_n) = c\alpha v_1+\cdots+ c \alpha v_n $ since $c\alpha \in R$ It is closed under scalar multiplication. And same for addition $( \alpha v_1+\cdots+\alpha v_n) + ( \beta v_1+\cdots+\beta v_n) = ( (\alpha + \beta) v_1+\cdots+(\alpha+\beta) v_n)$ and same $(\alpha + \beta) \in R$ using that $R$ is closed under addition. So it is valid subspace. Is this a good solution?
 A: You have a mistake as you've only considered linear combinations of the form $x = \alpha v_1 + \cdots + \alpha v_n$ - that is with the same coefficient through out and you haven't checked that $0 \in \text{span}(\{v_1,\ldots,v_n\})$. If, however, you do it how Dan mentions in the comments you must check that $\text{span}(\{v_1,\ldots,v_n\}) \neq \emptyset$. Also, your presentation could be better. Take a look at this which I hope is an improvement:

Let $V$ be a vector space. Prove that $W = \text{span}(\{v_1,\ldots,v_n\})$ is a subspace of $V$ where $v_1,\ldots,v_n \in V$. 

Proof: We have $0 \in W$ as $0 = 0 \cdot v_1 + \cdots + 0\cdot v_n$ is a linear combination of the vectors $v_1, \ldots, v_n$ and $0 \in K$. If $x,y \in W$ then they can be written as $x = \alpha_1 v_1 + \cdots + \alpha_n v_n$ and $y = \beta_1 v_1 + \cdots + \beta_n v_n$ for scalars $\alpha_1, \ldots, \alpha_n \in K$ and $\beta_1,\ldots,\beta_n \in K$, respectively. Note that $x + y = (\alpha_1 + \beta_1)v_1 + \cdots +  (\alpha_n + \beta_n)v_n$ which is a linear combination of the vectors $v_1, \ldots, v_n$ and scalars $\alpha_1 + \beta_1, \ldots, \alpha_n + \beta_n \in K$. Therefore, $x + y \in W$. In a similar fashion, you can show that if $x \in W$ then $cx \in W$ where $c \in K$. When you've done this, you can conclude that $W = \text{span}(\{v_1,\ldots,v_n\})$ is a subspace of $V$. 
Hope this helps!
A: You did well, though a few oversights in you argument/presentation. One oversight is the omission of explicitly noting that the zero vector is in $\operatorname{Span}\{v_i, \ldots, v_n\}$.
Indeed, given a vector space $V$ over a field $K,$ the span of a set $S$ of vectors is defined to be the intersection of all subspaces of $V$ that contain $S$ and this intersection is referred to as the subspace spanned by $S$ or by the vectors in $S$. 
Alternatively, we can define the span of $S$ as the set of all linear combinations of elements of $S$.
$$\operatorname{span}(S) = \left \{ {\sum_{i=1}^k \lambda_i v_i \mid k \in \mathbb{N}, v_i \in S, \lambda _i \in \bf{K}} \right \}.$$ 
And as you (almost) show, this linear combination of all the $v_i$ forms a subspace of $V$. (You can use the zero from the field as a scalar in the linear combination of all the $v_i$ to obtain the zero vector, which all subspaces must contain.)
