Proof that $\operatorname{span}(S_1\cap S_2) \subseteq \operatorname{span}(S_1) \cap \operatorname{span}(S_2)$ An exercise in my book says : 

Prove that $\operatorname{span}(S_1\cap S_2) \subseteq \operatorname{span}(S_1) \cap \operatorname{span}(S_2)$.
  Give an example in which $\operatorname{span}(S_1 \cap S_2)$ and $\operatorname{span}(S_1) \cap \operatorname{span}(S_2)$ are equal and one in which they are unequal.

Why are those two not necessarily equal?
Thanks a lot.
 A: I will write $[A]$ instead of $\operatorname{span}(A)$ for the sake of brevity.
Of course, the proofs depend on your definition of linear span. E.g., we can define the span of the set $A$ as the set of all linear combinations, i.e.
$$[A]=\{c_1\vec{a}_1+\dots+c_n\vec{a}_n; n\in\mathbb N; c_1,\dots,c_n\in F, a_1,\dots,a_n\in A\},$$
where $F$ is the fields we are working with.
Observation 1. $A\subseteq B$ implies $[A]\subseteq[B]$.
The proof is basically just observing that every linear combination of elements of $A$ is a linear combination of elements of $B$.
Observation 2. $[S_1\cap S_2] \subseteq [S_1]\cap[S_2]$.
From observation 1 we get $[S_1\cap S_2]\subseteq [S_1]$ and $[S_1\cap S_2]\subseteq [S_2]$, which yields $[S_1\cap S_2] \subseteq [S_1]\cap[S_2]$.
Example 3. We want to find an example where the inclusion is strict.
Let us try $S_1=\{e_1,e_2\}$ and $S_2=\{2e_1,e_3\}$, where $e_1=(1,0,0)$, $e_2=(0,1,0)$ and $e_3=(0,0,1)$.
We can see that: 


*

*$S_1\cap S_2=\emptyset$  and thus $[S_1\cap S_2]=\{0\}$

*$e_1\in[S_1]\cap [S_2] \subseteq [\{e_1\}]$ and $e_1\notin [S_1\cap S_2]$. So $[S_1\cap S_2]\subsetneqq [S_1]\cap[S_2]$ 
(In fact, we have $[S_1]\cap [S_2] = [\{e_1\}]$.)

