Splitting Fields This is a problem from Hungerford's book: 
Let $E$ be an intermediate field of extension $K\subset F$ and assume that $E=K(u_1, \cdots ,u_r)$ where $u_i$ are (some of the) roots of $f\in K[x]$. Then $F$ is a splitting field of $f$ over $K$ if and only if $F$ is a splitting field of $f$ over E. 
This is my attempt.  
Let $u_1, \cdots ,v_n$ be the roots of $f$. Then since $F$ is a splitting field of $f$ over $K$, $F=K(v_1, \cdots ,v_n)$ $\implies$ $F=E(v_1, \cdots ,v_n)$. So $F$ is a splitting field of $f$ over $E$ .  
Conversely, suppose $v_1, \cdots ,v_n$ are the roots of $f$ in $F$.  Then $F=E(v_1, \cdots ,v_n)=K(u_1, \cdots ,u_r,v_1, \cdots, v_n)$. Thus $F$ is a splitting field of $f$ over $K$.  
My Question: First I want to know if the proof above is right. The second part of the problem asks to extend the above to splitting fields of arbitrary  set of polynomials. This is where I'm lost. I'd like a little help on how to begin.
Thanks.  
Update:  I think I've got it now. I had some difficulty with the indexing. Thanks.  

$\left(\implies\right)$ Let $v_{1},\cdots,v_{n}$ be the roots of
$f$ in $F$. Since $F$ is a splitting field of $f$ over $K$, $F=K\left(v_{1,}\cdots,v_{n}\right)$
which implies that $F=E\left(v_{1},\cdots,v_{n}\right)$. So $F$
is a splitting field of $f$ over $E$.  
$\left(\Longleftarrow\right)$ Suppose $v_{1},\cdots,v_{n}$ are the
roots of $f$ in $F$. Then $F=E\left(v_{1},\cdots,v_{n}\right)$.
But each $u_{i}$ is a one of the $v_{j}$, so $F=E\left(v_{1},\cdots,v_{r}\right)=K\left(u_{1},\cdots,u_{r},v_{1},\cdots,v_{n}\right)$.  
Thus $F$ is a splitting field of $f$ over $K$.
