How to verify if a span of vectors is a subset of another span of vectors 
$\displaystyle b_1=\left\lbrace\frac{12}{9},\frac{12}{9},2\right\rbrace^T,b_2=\{-18,-18,21\}^T$ and $\displaystyle v_1=\{-1,-1,2\}^T,v_2=\{3,3,-3\}^T$. $b_1 \in \operatorname{Span}\{v_1,v_2\} \text{ and } b_2 \in \operatorname{Span}\{v_1,v_2\}$. Can we conclude that (Without performing further verification) $\operatorname{Span}\{b_1,b_2\} \subseteq \operatorname{Span}\{v_1,v_2\}$? What about $ \operatorname{Span}\{v_1,v_2\} \subseteq  \operatorname{Span}\{b_1,b_2\}$?

As the question has provided, both $b_1,b_2$ belongs to the span of $v_1,v_2$. However, the answer key only pointed out that we can only conclude $\operatorname{Span}\{b_1,b_2\} \subset \operatorname{Span}\{v_1,v_2\}$ but not $ \operatorname{Span}\{v_1,v_2\} \subset  \operatorname{Span}\{b_1,b_2\}$ without any further explanation.
Is the fact that both $b_1 \in \operatorname{Span}\{v_1,v_2\} \text{ and } b_2 \in \operatorname{Span}\{v_1,v_2\}$ implies that $\operatorname{Span}\{b_1,b_2\} \subseteq \operatorname{Span}\{v_1,v_2\}$ ?
More generally, as my question's title suggests, how do I verify if a Span of vectors is a subset of another span of vectors? (I am capable of verifying if a single vector belongs to a span of vectors) but I can't make any connections between the two
 A: If $b_1 \in \text{Span}(v_1, v_2) $ and $b_2 \in \text{Span}(v_1, v_2)$, then $\text{Span}(b_1, b_2) \subset \text{Span}(v_1, v_2)$.
Proof:
$b_1 \in \text{Span}(v_1, v_2) $ $\implies $ $b_1 = k_1 v_1 + k_2 v_2$. Similarly, $b_2 \in \text{Span}(v_1, v_2)$ $\implies $ $b_2 = c_1 v_1 + c_2 v_2$.
Hence, an arbitrary vector $w \in \text{Span}(b_1, b_2)$ satisfies
\begin{align}
  w &= m_1 b_1 + m_2 b_2\\
    &= (m_1 k_1 + m_2 c_1) v_1 + (m_1 k_2 + m_2 c_2 ) v_2 
\end{align}
This implies $w \in \text{Span}(v_1, v_2)$. Hence, $\text{Span}(b_1, b_2) \subset \text{Span}(v_1, v_2)$.
End Proof
If $b_1 \in \text{Span}(v_1, v_2) $ and $b_2 \in \text{Span}(v_1, v_2)$, then $\text{Span}(v_1, v_2) \subset \text{Span}(b_1, b_2)$ is in general False.
Counterexample
Take $b_1 = b_2 = v_1$. Then, Span$(b_1, b_2)$ = Span$(v_1)$. Hence,  $\text{Span}(v_1, v_2) \nsubseteq \text{Span}(b_1, b_2)$
Response to comments:
If $b_1 \in \text{Span}(v_1, v_2) $ and $b_2 \in \text{Span}(v_1, v_2)$, then $\text{Span}(b_1, b_2) \subset \text{Span}(v_1, v_2)$.
Now if $\text{Span}(b_1, b_2) \subset \text{Span}(v_1, v_2)$  and dim(Span($b_1, b_2$)) $=$ dim(Span($v_1, v_2$)), then indeed Span($b_1, b_2$) $=$ Span($v_1, v_2$). This follows from a general rule that any subspace $A$ of a vector space $V$ satisfying dim$(A) = $ dim($V$) implies that $A = V$.
A: As @ironX & @pietro said, and also by definition of vector spaces, they contain the $Span$ of any subset of themselves. But for inverse, also as they said truly, you can't conclude it. Of course, if you are curious to find such condition:
If you have $\dim(Span\{ v_1, v_2 \}) \leqslant \dim(Span\{ b_1, b_2 \})$, beside $Span\{ b_1, b_2 \} \subseteq Span\{ v_1, v_2 \}$, then you can conclude the inverse: $Span\{ v_1, v_2 \} \subseteq Span\{ b_1, b_2 \}$ and so they become equal with equal dimension.
In some cases investigating the independency of vectors can help much.
A: Indeed you cannot say that $Span\{v_1,v_2\} \subseteq Span\{b_1,b_2\}$ in general. To see that consider the trivial case where:
$v_1=\begin{pmatrix} 1 \\ 0 \end{pmatrix}$, $v_2=\begin{pmatrix} 0 \\ 1 \end{pmatrix}$
and $b_1=b_2=\begin{pmatrix} 1 \\ 0 \end{pmatrix}$.
It is clear that $Span\{b_1,b_2\}=Span\{b_1\} \subseteq Span\{v_1,v_2\}$ but not the contrary (you cannot generate $v_2$ using $b_1$).
Now, about your question "Is the fact that both $b_1 \in Span\{v_1,v_2\} \text{ and } b_2 \in Span\{v_1,v_2\}$ implies that $Span\{b_1,b_2\} \subseteq Span\{v_1,v_2\}$ ?" The answer is yes.
To see that, go back to the definition of $Span\{v_1,...,v_n\}$ : it is the set of vectors that can be genereated by a linear combination of $(v_1,...,v_n)$.
In your case, as $b_1$ and $b_2$ can be computed as linear combinations of $v_1$ and $v_2$, thus any linear combination of $b_1$ and $b_2$ can be computed as linear combinations of $v_1$ and $v_2$. Hence the result.
Now a general method to see if a Span of vectors $b_1,...,b_k$ is a subset of another span of vectors $v_1,...,v_k$ is to check if for all i in 1,...,k $b_i$ is a linear combination of $v_1,...,v_k$.
Hope it helps you!
