Question about the proof of: $E/k$ is algebraic of characteristic zero. Every $f(x)\in k[x]$ has a root in $E$. Then $E$ is algebraically closed.

Question

Question

Let $k\subset E$ be an algebraic extension of fields of characteristic zero. Assume that every non-constant polynomial $f(x)\in k[x]$ has a root in $E$. Show that $E$ is algebraically closed.

Answer

I was following one of the answers given in this link, namely the last one. The answers follows like:

"First take $f\in k[x]$. Then as characteristic is zero $Spl(f)=k(\alpha)$. Then $m_{\alpha,k}(x)$, minimal polynomial of $\alpha$ in $k[x]$, has a root in $E$. Thus, $E$ contains a copy of the splitting field of $f$."

I didn't understand how the last sentence holds. Thanks in advance...

Also related to this question.