Sum of subspaces equal to span of union of subspaces 
Show that $\sum Wj = Span(\cup
W_j )$ for a family of subspaces ${Wj}_{j\in J}$ of a vector space V .

So I am trying to use induction to prove the result of a family of subspaces here but am stuck on the induction step. Please see below.
Base Case:
We use double inclusion/containment to show $W_1+W_2=Span(W_1\cup W_2)$.
If $x\in W_1+W_2$ then $x=w_1+w_2$ for $w_i\in W_i$. However this immediately implies $x\in Span(W_1\cup W_2)$ because $x$ is a sum of elements from $W_1\cup _2$. This gives the first inclusion.
If $x\in Span(W_1\cup W_2)$, then we can write $x$ as a linear combination of elements from $W_1\cup W_2$, say
$$x=a_1u_1+a_2u_2+\cdots+a_ku_k+b_1w_1+b_2w_2+\cdots+b_jw_j$$
where the $a_i, b_i\in\mathbb{R}$, $u_i\in W_1$, and $w_i\in W_2$.  But then we observe that
$$a_1u_1+a_2u_2+\cdots+a_ku_k\in W_1$$
and
$$b_1w_1+b_2w_2+\cdots+b_jw_j\in W_2.$$
Then $x\in W_1+W_2$ because $a_1u_1+a_2u_2+\cdots+a_ku_k\in W_1$
and
$b_1w_1+b_2w_2+\cdots+b_jw_j\in W_2$.
This gives the other inclusion.
Induction step:
Given $W_1+\ldots+W_n=Span(W_1\cup \ldots \cup W_n)$, we need to show $W_1+\ldots+W_{n+1}=Span(W_1\cup \ldots \cup W_{n+1})$.
This is where I am not sure how to continue; is induction even a good solution here or I should be able to directly get to the result without using a base case?
 A: We can show this by proving that $x\in\sum W_j \iff x\in\operatorname{span}(\cup W_j)$.
Let $x\in\sum W_j.$ Then, $x = \sum w_j $ where $w_j\in W_j$ for every $j$. Thus, every $w_j\in\cup W_j$ which means $\sum w_j \in \operatorname {span}(\cup W_j)$. Thus, $x\in\operatorname {span}(\cup W_j)$.
Conversely, let $x \in \operatorname{span}(\cup W_j)$. Then, $x=\sum v_i$, where each $v_i\in\cup{W_j}$. Thus, each $v_i$ is in at least one of the $W_j$'s. Now, if two or more $v_i$'s are in the same space $W_j$, we can combine them linearly to get a new vector which is in $W_j$. Thus, $x=\sum u_j$, where each $u_j\in W_j$ (some of them may be zero also). Thus, $x\in \sum W_j$.
A: By definition, $\sum W_j$ is the set of finite sums $v_1 + \cdots + v_n$ where each $v_i$ is in some $W_j$.
By definition, span$(\bigcup_j W_j)$ is the set of finite linear combinations $\lambda_1 v_1 + \cdots \lambda_n v_n$ where each $\lambda_i$ is a scalar and each $v_i$ is in some $W_j$.
This subsumes the previous case by taking $\lambda_i = 1$ for each $i$.
Conversely, if $v_i \in W_{j_i}$ then $\lambda_i v_i \in W_{j_i}$ as well.
