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We fix a positive prime number $p$ and a field $k$ such that $p\neq \text{Char}\left (k\right )$ and $k$ has a primitive $p$-th root of unity.

We say that a finite and separable extension $E/k$ is $p$-solvable if the Galois group of its normal closure is a $p$-group.

Let $\overline{k_p}$ be the union of all fields which are $p$-solvable over $k$. We call $\overline{k_p}$ the $p$-closure of $k$.

It is easy to show that if $\alpha \in \overline{k}$ and $\alpha^p\in k$ then $\alpha \in \overline{k_p}$, moreover, if $\alpha \in \overline{k}$ and $\alpha^p\in \overline{k_p}$ then $\alpha\in\overline{k_p}$.

Therefore $\overline{k_p}$ is closed under $p$-th roots. But why is $\overline{k_p}$ the minimum extension satisfying this property?

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Your $p$-closure $k_p$ (I drop the over bar) is a normal pro-$p$-extension of $k$, i.e. $Gal(k_p/k)$ is a pro-$p$-group (which could be finite). Let $k_{pcl}$ be the subextension of $k_p$ which is closed under the operation $\sqrt [p].$, and put $G=Gal(k_p/k_{pcl})$. Because $G$ is a pro-$p$-group, the equality $k_p=k_{pcl}$ is equivalent to the triviality of $G$ or, by the Burnside basis theorem, to the triviality of the quotient $G^* = G/adh (G^p [G, G])$ (see e.g. Serre's "Galois Cohomology", §4, prop. 25). In other terms, we must show that $k_{pcl}$ coincides with the subfield $L$ of $k_p$ which is the composite of all the cyclic extensions of degree $p$ of $k_{pcl}$ contained in $k_p$. But $k_{pcl}$ has characteristic prime to $p$ and contains a primitive $p$-th root of $1$, so we can apply Kummer's theory, and the closedness of $k_{pcl}$ w.r.t. $\sqrt [p].$ shows what we want.

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    I couldn't understand since I don't know about the theory you used. In class, we proved that $E/k$ is $p$-solvable iff there exists a field $E\subset F$ and a tower $k=E_0\subset E_1\subset \cdots \subset E_m$ such that $E_{i+1}=E_i\left (\alpha_i\right )$ with $\alpha_i^p\in E_i$, then we proved what I said about $\overline{k_p}$. Finally, our professor wrote that $\overline{k_p}$ was the minimum extension satisfying closedness under $p$-th roots as an immediate corollary, but I cannot see how it is implied. – 2017-02-15
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    I suspect it must be the Burnside basis theorem that you don't understand. It is not indispensable here, it's just a convenient way to deal with "infinite p-extensions" (= pro-p-extensions). You can argue in $k_p/k_pcl$ as follows. Consider any finite p-extension $E/k_pcl$ contained in $k_p$ and the chain $E_i$ that you wrote down. Kummer theory and the closedness of $k_pcl$ show that $E_1 = k_pcl$, and so on inductively, so there is no non trivial such chain. Intuitively this suffices to prove $ k_p = k_pcl$, but rigorously, you need a topological argument.. which is BB's theorem. – 2017-02-16