If $\operatorname{sp}(A) \cup \operatorname{sp}(B)=\operatorname{sp}(A\cup B) \Rightarrow A\cup B$ is linearly dependent $\newcommand{\sp}{\operatorname{sp}}$
Let $V$ be a vector space over $F$ field, and let $A,B$ be two different, disjoint, non empty sets of vectors from $V$.
If $\sp(A) \cup \sp(B)=\sp(A\cup B) \Rightarrow A\cup B$ is linearly dependent
I've started by saying that if $\sp(A) \cup \sp(B)=\sp(A\cup B)$
then $\sp(A) \subseteq \sp(B)$ or $\sp(B)\subseteq \sp(A)$.
Thus if we assume, WLOG, that $\sp(A) \subseteq \sp(B)$ and we take $v_1 \in A$ and multiply it by scalars $\alpha_1,\ldots,\alpha_k$ we'll get a vector from $B$.
Thus $ A\cup B$ is linearly dependent.
I feel like this proof is not good enough, where is it failing?
 A: We use contraposition at this proof:
suppose $A\cup B$ be linear independence then elements of A as $\{\alpha_1,..,\alpha_n\}$ with elements of B $\{\beta_1,...\beta_m\}$ will be linear inedependence so $\alpha_1+....+\alpha_n+\beta_1+....\beta_m \in span(A\cup B)$ but it is not belong to span(A)$\cup$span(B)
A: Here is a direct proof:
To prove that $A \cup B$ is linearly dependent I must show that there exists a finite number of distinct vectors $w_1, \ldots,w_n$ in $A \cup B$ and scalars $c_1,\ldots,c_n$ in $F$, not all zero, such that 
$$ c_1w_1 + \dots + c_nw_n = 0.$$
As already mentioned, span$(A) \cup$ span$(B) =$ span$(A \cup B)$ implies span$(A) \subset$ span$(B)$ or span$(B) \subset$ span$(A)$. I'll assume the former. 
Let $x \neq 0 \in$ span$(A)$. Therefore, we can write $x = a_1v_1 + \dots + a_sv_s$ for some $v_1,\dots,v_s$ in $A$ and for scalars $a_1,\dots, a_s$ in $F$. By the inclusion above, we have that $x \in$ span$(B)$ and similarly, $x = b_1u_1 + \dots + b_tu_t$ for $u_1,\ldots,u_t \in B$ and $b_1,\ldots, b_t \in F$. Note then that we have
$$0 = x - x = (a_1v_1 + \dots + a_sv_s) - (b_1u_1 + \dots + b_tu_t)$$
Which is a linear combination of distinct vectors (as $A$ and $B$ were disjoint) of $A \cup B$ and the scalars $a_1,\ldots,a_s, b_1, \ldots, b_t$ are not all zero as $x$ was non-zero. Therefore, $A \cup B$ is linearly dependent. 
A: Your idea is good, but not expressed in the best way.
You can assume, as you did, $\def\sp{\mathrm{sp}}\sp(A)\subseteq\sp(B)$. If the only vector in $A$ is $0$, you have nothing to prove, because any set containing the zero vector is linearly dependent. Otherwise, take $v\in A$, $v\ne0$. Then you can write
$$
v=\alpha_1w_1+\alpha_2+\dots+\alpha_nw_n
$$
for suitable $w_1,w_2,\dots,w_n\in B$ and so $A\cup B$ is not linearly independent.
