If $S'$ is a finite subset of $ \text{Span}(S)$ then $\text{Span}(S') \subset \text{Span}(S)$.
Let $\displaystyle \text{span}(S')= \left\{\sum_{1 \le k \le m} a'_k v'_k: a'_k \in F, v'_k \in S'\right\}$, and $\displaystyle \text{span}(S)= \left\{\sum_{1 \le i \le n} a_i v_i: a_i \in F, v_i \in S \right\}$.
Since $S' \subset \text{Span}(S)$ then we can write every $v'_k \in S$ in the form
$$v'_k = \displaystyle \sum_{j \in \left\{1, \cdots, n \right\}}a_jv_j $$
Multiplying by $a'_k$ and summing over $k$ we have:
$$ \displaystyle \sum_{1 \le k \le m} a'_kv'_k = \sum_{1 \le k \le m} \sum_{j \in \left\{1, \cdots, n \right\}}a'_ka_jv_j = \sum_{j \in \left\{1, \cdots, n \right\}}(\lambda a_j) v_j $$
Where $\lambda$ is the (scalar) sum of the $a'_k$. Any element in $\text{span}(S')$ can be written in the form of the left hand side by definition, and the right hand side is a subset of $\text{span}(S)$ hence $\text{Span}(S') \subset \text{Span}(S)$.
Does that make sense?
That approach is correct, but there is no reason for you to write$$\sum_{j\in\{1,\ldots,n\}}\text{ instead of }\sum_{j=1}^n.$$And it is not correct to state that “the right hand side is a subset of $\operatorname{span}(S)$”; the right hand side is an element of $\operatorname{span}(S)$.