$S_1 \subset S_2$. To show, $Span(S_1) \subset Span(S_2)$ Prove that if $S_{1} \subset S_{2}$, then $Span(S_{1}) \subset Span(S_{2})$
Approach: Suppose $S_{1} \subset S_{2}$
Let $x \in S_{1}$, then by definition of a subset, $x \in S_{2}$
All possible linear combinations of $x$ are $cx$ with $c \in \mathbb{R}$
So $cx \in Span(S_{1})$
But $x$ is also in $S_{2}$, therefore $Span(S_{2})$ contains all linear combinations of $x$ as well.
So $cx \in Span(S_{2})$
We have $cx \in Span(S_{1})$, then $cx \in Span({S2})$  $\forall c \in \mathbb{R}$
We have shown: if $S_{1} \subset S_{2} \Rightarrow Span(S_{1}) \subset Span(S_{2})$
Is my approach to the question correct? If so, is this enough to answer the question or have I missed something?

No, that is not a valid approach.
In order to show that $\mathit{Span}(S_1)\subseteq \mathit{Span}(S_2)$ you need to start with an arbitrary element of $\mathit{Span}(S_1)$ and show that it is also an element of $\mathit{Span}(S_2)$.
You have shown this only for elements of $\mathit{Span}(S_1)$ that happen to have the form $cx$ for some $c$ and and $x\in S_1$, but there will generally be elements of $\mathit{Span}(S_1)$ that do not have this simple form.
