How to show $S_1\subset W_1$ and $S_2\subset W_2$ are independent $\implies$ $S_1\cup S_2$ is independent based on the following assumption? Let $W_1$ and $W_2$ be subspaces of vector space $V$ satisfying $W_1\cap W_2=\{0\}$ ,how to show $S_1\subset W_1$ and $S_2\subset W_2$ are linearly independent $\implies$ $S_1\cup S_2$ is linearly independent ?
Here is my proof:
let $v\in $ $S_1\cup S_2$, $s_1\in S_1,\ s_2\in S_2$. 
Express $s_1=a_1x_1+...+a_ix_i$ for $x_j$ from $W_1$ and $s_2=b_1y_1+...+b_ny_n$ for $y_l$ from $W_2$.
Since both linear combinations are linearly independent, there are no dependent vectors in each one. In addition, because $W_1\cap W_2=\{0\}$, the two sets are linearly independent of each other. 
Express $v=a_1x_1+...+a_ix_i+ b_1y_1+...+b_ny_n$, $S_1\cup S_2$ is linearly independent.
Could someone correct my proof
 A: Usually when one checks for linear independence of a set of vectors, one takes a linear combination of finitely many vectors from the set, equates them to zero and shows that each scalar in the combination is necessarily zero.
Note that since $S_1\subseteq W_1$ and $S_2\subseteq W_2$ then $S_1\cap S_2\subseteq W_1\cap W_2=\{0\}$. Since $S_1$ and $S_2$ are linearly independent, $0$ does not belong to any of them. Thus $S_1\cap S_2=\emptyset$.
Now let $v_1,\cdots, v_n\in S_1\cup S_2$ and suppose $a_1v_1+\cdots +a_nv_n=0$. Each $v_i$ is in either $S_1$ or $S_2$ but not both. WLOG, assume $v_1,\cdots, v_j\in S_1\subseteq W_1$ and $v_{j+1},\cdots,v_n\in S_2\subseteq W_2$. Then $$a_1v_1+\cdots+a_jv_j=-(a_{j+1}v_{j+1}+\cdots+a_nv_n)$$
But the LHS of the above equation is in $W_1$ and the RHS is in $W_2$ and $W_1\cap W_2=\{0\}$. This means that $$a_1v_1+\cdots+a_jv_j=0,$$ $$a_{j+1}v_{j+1}+\cdots+a_nv_n=0$$
Bot since $S_1$ and $S_2$ are both linearly independent sets, we have that each $a_i=0.$ $\square$
