Sum of two basis in same vector space is again forms a basis under $|I|+|J|=\operatorname{dim}(V)=n$. Let $A=\{\alpha_1, \dots, \alpha_n\}$ and $B=\{\beta_1, \dots, \beta_n\}$ be two bases of a vector space $V$.
I want to

Show that for any $I\subset A$, there exists $J\subset B$ such that $|I|+|J| =n$ and $I\cup J$ is a basis of $V$.


Since $A,B$ are basis, one thing for sure is the element in $I,J$ are linearly independent, respectively.  But I am not sure their union with $|I|+|J|=n$ be a basis.
 A: First note that your observation that elements in $I$ and $J$ are linearly independent is generally not true apriori. You have to construct $J$ from $I$ or vice versa to make it so. For example take $A=\{(1,0,0),(0,1,0),(0,0,1)\}$ and $B=\{(1,0,0,),(1,0,1),(1,1,0)\}$. Then $A$ and $B$ are both bases for $\mathbb{R^{3}}$ , but for the choice of $I=\{(1,0,0)\}$ and $J=\{(1,0,0),(1,0,1)\}$ you cannot conclude that $I\cup J$ is a basis.
For easier typing let $A=\{v_{1},v_{2},...,v_{n}\}$ and $B=\{w_{1},w_{2},...,w_{n}\}$ be the respective bases for $V$.
Let $|I|=k<n$ . Without loss of Generality assume $I=\{v_{1},v_{2},...,v_{k}\}$. It is clear that $I$ is a linearly independent set.
(We can assume this because once we have those k elements we can just re-index them so that they are the first $k$ vectors in our ordered basis $A$. )
Now consider $w_{1}\in B$. If $w_{1}\in \text{span}(I)$ then discard it and proceed to $w_{2}$. Repeat the same. This process must stop before we exhaust all $w_{i}\in B$ because if it were not the case, we would have all of $w_{i}$'s to be in the span of $I$ and since $B$ is a basis , $V=span(B)\subset span(I)\implies I$ is a basis. But this contradicts that the dimension of $V$ is $n$.
So let the process stop at $w_{m}$ . Now you reindex $w_{i}$ 's such that $w_{m}=w'_{1}$ and make $B=\{w'_{1},w^{(1)}_{2},...,w^{(1)}_{n}\}$.
Now you consider the set $I\cup\{w'_{1}\}=I^{1}$.
Now repeat the same process with $I^1$ in place of $I$ and you will again get a $w'_{2}$.
In this step $B=\{w'_{1},w'_{2},w^{(2)}_{3},...,w^{(2)}_{n}\}$.
Repeat this process inductively untill you get $n-k$ vectors $\{w'_{1},w'_{2},...,w'_{n-k}\}$ and call this set as $J$. That is we arrive at $I^{n-k} = I\cup J$.
Notice that you are always guaranteed that you will get $n-k$ vectors because if not you will arrive at a linearly independent spanning set of dimension less than $n$.
Now once we have $I^{n-k}$, it's cardinality is $n$. And it is linearly independent as vectors in $I$ are linearly independent and members in $J$ are linearly independent and members of $I$ do not contain members of $J$ in their span. (Basically it is linearly independent by construction.)
Thus we have that $I\cup J$ is a linearly independent set of $n$ vectors and hence a basis.
