Sum of two subspaces is equal to the span of their union

If $x\in V_1+V_2$ then $x=v_1+v_2$ for $v_i\in V_i$. However this immediately implies $x\in Span(V_1\cup V_2)$ because $x$ is a sum of elements from $V_1\cup V_2$. This gives the first inclusion.

If $x\in Span(V_1\cup V_2)$, then I can write $x$ as a sum of elements from $V_1\cup V_2$, say $u_1+w_1+u_2+w_2+\dots+u_k+w_k$, where $u_i\in V_1$ and $w_i\in V_2$. Then I can group these so

$$x=(u_1+\dots+u_k)+(w_1+\dots+w_k)\in V_1+V_2$$

because $u_1+\dots+u_k\in V_1$, and $w_1+\dots+w_k\in V_2$. This gives the other inclusion and you are done.

I think you're on the right track:

To demonstrate this $V_{1}+V_{2} \subset Span(V_{1} \cup V_{2})$ you need to pick a vector $w$ in $V_{1}+V_{2}$ and prove that is in the $Span(V_{1} \cup V_{2})$ writing it as a linear combination of vector of $V_{1}$ and $V_{2}$ (indeed is trivially from the definition).

The converse $Span(V_{1} \cup V_{2}) \subset V_{1}+V_{2}$ is similar, i.e. pick a vector in the span write it in a base of $Span(V_{1} \cup V_{2})$ then rearrange the term to show that is in $V_{1}+V_{2}$.

This is the kind of question that is hard because of how trivial it is. Just rewrite the RHS and the LHS in terms of their definitions and you basically have a proof.