Let u4 be a vector that is not a linear combination of {u1,u2,u3,}. Select the best statement.

A) span{u1, u2, u3} = span{u1, u2, u3, u4}

B) There is no obvious relationship between span{u1, u2, u3} and span{u1, u2, u3, u4}

C) span{u1, u2, u3} is a proper subset of span{u1, u2, u3, u4}

D) We only know that span{u1, u2, u3} ⊆ span{u1, u2, u3, u4}

E) span{u1, u2, u3} = span{u1, u2, u3, u4} when none of  {u1, u2, u3, }  is a linear combination of the others. 

F) We only know that span{u1, u2, u3, u4} = span{u1, u2, u3}

G) None of the above. 

I have tried B and G thinking that there is not an explicit relationship between the two sets, however, both those choices are incorrect.
For any $u_1, u_2, u_3, u_4$, we always have
$$\text{span}\{u_1, u_2, u_3\} \subseteq \text{span}\{u_1, u_2, u_3, u_4\}.$$
This is because any linear combination of $u_1, u_2, u_3$ is automatically a linear combination of $u_1, u_2, u_3, u_4$. (Explicitly, $c_1 u_1 + c_2 u_2 + c_3 u_3 = c_1 u_1 + c_2 u_2 + c_3 u_3 + 0 \cdot u_4$.

Now the question is whether this containment is strict containment $\subsetneq$ or equality $=$. If $u_4$ is not a linear combination of $u_1, u_2, u_3$, then it does not lie in $\text{span}\{u_1, u_2, u_3\}$. Therefore the containment must be strict $\subsetneq$.
Assuming there is no additional information you have not told us, D is correct. 

A) "$span(x_1, x_2, x_3)= span(x_1, x_2, x_3, x_4)$" is true only if $x_4$ is a linear combination of $x_1$, $x_2$, and $x_3$.

B) It's not clear what "obvious relationship" means here but we can certainly say that 
$span(x_1, x_2, x_3)$ is a subspace of $span(x_1, x_2, x_3, x_4)$.  I would call that an "obvious relationship"!

C)  I just said it was a subspace but it is not necessarily a "proper" subspace.

D)  This is what being a subspace means.

E)  Again not necessarily true because $x_4$ is not necessarily a linear combination of the others.

F)  This is just a restatement of (A) and (E)!