i think my proof is correct but i'm not sure. it begins on the next line. $u$ and $v$ are distinct vectors.
$\{u,v\}$ is a basis, so:
$a_1u+a_2v=0, a_1=a_2=0$ by definition
consider the linear combination of $\{u+v,au\}$. let $c \in F$.
$c(u+v)+au=0$
$cu+cv+au=0$
$(c+a)u+cv=0$
by above linear combination for $\{u,v\}$, $c+a=a_1=c=a_2=0$, so $\{u+v,au\}$ must be linearly independent. and by properties of $\{u,v\}$, $\{u+v,au\}$ must also be generating. therefore this set is a basis for $V$.
consider linear combination of $\{au,bv\}$. let $p,q \in F$.
$p(au)+q(bv)=0$
$(pa)u+(qb)v=0$
by the linear combination for $\{u,v\}$, the linear combination of $\{au,bv\}$ is equivalent, which implies $pa=a_1=qb=a_2=0$, so $\{au,bv\}$ must be linearly independent. it also must be a generating set by the characteristics of $\{u,v\}$. therefore $\{au,bv\}$ is also a basis.