$\operatorname{Span}(A \cap B) \subseteq \operatorname{Span}(A) \cap \operatorname{Span}(B)$ So I'm trying to prove that $$\operatorname{Span}(A\cap B)  \subseteq \operatorname{Span}(A) \cap \operatorname{Span}(B)$$
From the definition I have that $$x\in \operatorname{Span}(A)$$ and also $$x\in \operatorname{Span}(B)$$
And from the definition of Span, i take $$\operatorname{Span}(A)= \left\{x_{1},x_{2},...,x_{k}\right\} $$
 $$\operatorname{Span}(B)= \left\{y_{1},y_{2},...,y_{k}\right\} $$
How to continue this?
 A: You seem to be confusing ‘span of a subset’ of a vector space and  ‘set of generators’. 
Hint:
Just prove that, if $X\subseteq Y$, then $\;\operatorname{Span}(X)\subseteq \operatorname{Span}(Y)$.
Added (kind suggestion of @egreg):
The following example shows we can have a proper inclusion:
Take a non-zero vector $v$, $A=\{v\}$, $B=\{-v\}$ (we suppose the characteristic of the field is $\ne 2$). Then $\operatorname{Span}A=\operatorname{Span}B$, but $\;A\cap B=\varnothing$, so $\;\operatorname{Span}\varnothing =\{0\}$.
A: You can solve this by element chasing: take an $x\in\mathrm{span}(A\cap B)$. Thus $x=\sum_{i=1}^kc_i v_i$, for coefficients $c_i$ in $\mathbb{K}$ (this is the field where the scalars come from, could be $\mathbb{R}$ for example) and $v_i\in A\cap B$, for all $1\leq i\leq k$. Thus $v_i\in A$ and $v_i\in B$ for all $1\leq i\leq k$, and thus $x\in\mathrm{span}(A)$ (being a linear combination of vectors in $A$). Similarly, $x\in\mathrm{span}(B)$, thus $x\in\mathrm{span}(A)\cap\mathrm{span}(B)$. Then, since $x\in\mathrm{span}(A\cap B)$ was arbitrary, we conclude $\mathrm{span}(A\cap B)\subset\mathrm{span}(A)\cap\mathrm{span}(B)$.
A: Let $v\in Span(A\cap B)$ so you can write $v=c_1x_1+ \dots +c_nx_x$ where $c_1 \dots c_n \in F$ F is any field.
And the vectors $x_1 \dots x_n \in A$ but also, by definition of intersection $x_1 \dots x_n \in B$
So we have that 
$v\in Span(A)$ and $v\in Span(B)$ hence $v\in Span(A)\cap Span(B)$ because $v$ is expressed as a linear combination of elements of A and B.
Concluding $\mathrm{span}(A\cap B)\subset\mathrm{span}(A)\cap\mathrm{span}(B)$.
