$\implies$ : Simply take $S = K$.

$\impliedby$ : We divide it into two cases.

Case 1: $[K : F] < \infty$. Use induction on $[K : F]$. The inertial case $[K : F] = 1$ is obviously true. Assume the statement holds for all field extensions of degree $\leq n - 1$. Suppose $[K : F] = n, K = F(S)$ and every element in $S$ is separable over $F$. Take any $\alpha \in S \setminus F$, $\alpha$ is separable over $F$ $\implies$ $\#\{ F\text{-embeddings }\varphi : F(\alpha) \to \overline{F} \} = [F(\alpha) : F]$ by Proposition 17010. Moreover, by induction hypothesis, $[K : F(\alpha)] < [K : F] = n$ (bc $\alpha \not\in F$), it follows that $K / F(\alpha)$ is separable. So by Proposition 1707, $\#\{ F(\alpha)\text{-embeddings} \varphi : K \to F(\alpha) \} = [K : F(\alpha)]$. Thus by Proposition 17010, $$ \begin{align*} &\#\{ F\text{-embeddings } \varphi : K \to \overline{F} \} \\ = &\#\{ F\text{-embeddings } \varphi : F(\alpha) \to \overline{F} \} \cdot \#\{ F(\alpha)\text{-embeddings } \varphi : K \to \overline{F(\alpha)} = \overline{F} \} \\ = &[F(\alpha) : F][K : F(\alpha)] \\ = &[K : F]. \end{align*} $$ So again, by Proposition 1707, $K / F$ is separable.

Case 2: $[K : F] = \infty$. Since $K = F(S)$, given any $\beta \in K$, there exists $\alpha_{1}, \dots, \alpha_{n} \in S$ such that $\beta \in F(\alpha_{1}, \dots, \alpha_{n})$. Now Case 1 $\implies$ $F(\alpha_{1}, \dots, \alpha_{n}) / F$ is separable, hence $\beta$ is separable over $F$. Therefore, $K / F$ is separable.

1. Let $S \subseteq \overline{\mathbb{F}}_{p}$ be the set of roots of $x^{p^{n}} - x$. We first show that $S$ is a subfield of $\overline{\mathbb{F}}_{p}$, then we show that $\lvert S \rvert = p^{n} = \deg (x^{p^{n}} - x)$, this suffices to show that $x^{p^{n}} - x$ is separable.

   First, $0 \in S$. Given $\alpha, \beta \in S$, we have $(\alpha - \beta)^{p^{n}} - (\alpha - \beta) = \alpha^{p^{n}} - \beta^{p^{n}} - (\alpha - \beta) = 0$. Hence $\alpha - \beta \in S \implies (S, +)$ is a subgroup of $(\overline{\mathbb{F}}_{p}, +)$. Next, $(\alpha\beta)^{p^{n}} - \alpha\beta = \alpha^{p^{n}}\beta^{p^{n}} - \alpha\beta = 0 \implies \alpha\beta \in S$ and $(\alpha ^{-1})^{p^{n}} - \alpha ^{-1} = (\alpha^{p^{n}})^{-1} - \alpha ^{-1} = 0 \implies \alpha ^{-1} \in S$ shows that $S$ is a subfield of $\overline{\mathbb{F}}_{p}$. Moreover, $\mathbb{F}_{p^{n}} = \mathbb{F}_{p}[S] = S$. This is because any $\alpha \in \mathbb{F}_{p}$ satisfies $\alpha^{p^{n}} - \alpha = \alpha^{p^{n} - 1}\alpha - \alpha = 1 \cdot \alpha - \alpha = 0$ by Fermat’s little theorem. Hence $\alpha \in S$ $\implies$ $\mathbb{F}_{p}[S] = S$.

   Now since $(x^{p^{n}} - x)' = p^{n}x^{p^{n} - 1} - 1 = -1$, we have $\gcd(x^{p^{n}} - x, (x^{p^{n}} - x)') = 1$. By Proposition 1704, $x^{p^{n}} - x$ is separable over $\mathbb{F}_{p}$, hence $\lvert S \rvert = p^{n}$.

   Then we prove the uniqueness of $\mathbb{F}_{p^{n}}$. Suppose that $F$ is a subfield of $\overline{\mathbb{F}}_{p}$ with $\lvert F \rvert = p^{n}$. Then the multiplicative group $F^{\times}$ is abelian of order $p^{n} - 1$. By Lagrange’s Theorem, we know that $\alpha^{p^{n} - 1} = 1$ for all $\alpha \in F^{\times}$. Hence every element in $F$ is a root of $x^{p^{n}} - x$, hence $F \subseteq \mathbb{F}_{p^{n}}$. Since they have the same cardinality, $F = \mathbb{F}_{p^{n}}$.

2. Let $F_{0}$ be $F$’s prime field. By the definition of $[F : F_{0}]$, which is $\dim_{F_{0}} F$, we have $\lvert F \rvert = \lvert F_{0} \rvert^{[F:F_{0}]}$. Hence $\lvert F_{0} \rvert$ is a prime that divides $p^{n}$ $\implies$ $\lvert F_{0} \rvert = p \implies F_{0} \cong \mathbb{F}_{p}$. Let $\varphi : F_{0} \to \overline{\mathbb{F}}_{p}$ be defined by $F_{0} \xrightarrow{\cong} \mathbb{F}_{p} \hookrightarrow \overline{\mathbb{F}}_{p}$. By [Proposition 1603], $\varphi$ extends to a field embedding $\psi : F \to \overline{\mathbb{F}}_{p}$. Then $\mathrm{Im}\psi \cong F$ is a subfield of $\overline{\mathbb{F}}_{p}$ and $\lvert \mathrm{Im}\psi \rvert = \lvert F \rvert = p^{n}$. By (1), we know that $\mathrm{Im}\psi = \mathbb{F}_{p^{n}}$. Therefore, $F \cong \mathbb{F}_{p^{n}}$.

1. Let $F$ be a field of characteristic $0$. Given finite field extension $K / F$ and $\alpha \in K$, let $m_{\alpha, F}(x) \in F[x]$ be the minimal polynomial of $\alpha$ over $F$. Then $m_{\alpha, F}(x)$ is irreducible. By (3) of [Proposition 1704], $m_{\alpha, F}(x)$ is separable. Hence $\alpha$ is separable over $F$. So $K / F$ is separable $\implies$ $F$ is perfect.
2. Let $F$ be a finite field with $\operatorname{ch}(F) = p$ and $\lvert F \rvert = p^{n}$ for some integer $n$. By the proof of Theorem 1801 ,$\alpha = \alpha^{p^{n}} = (\alpha^{p^{n-1}})^{p}$ for every $\alpha \in F$. In particular, every element in $F$ is a $p$-th power in $F$. By (3), $F$ is perfect.
3. Suppose that $F = F^{p}$. Let $K / F$ be a finite extension and $\alpha \in K$. Let $m_{\alpha, F}(x)$ be the minimal polynomial of $\alpha$ over $F$. We show that $m_{\alpha, F}(x)$ is separable by contradiction. Suppose the opposite. Then by (2) of Proposition 1704, $\gcd \left( m_{\alpha, F}(x), m_{\alpha, F}'(x) \right) \neq 1$, then $\gcd \left( m_{\alpha, F}(x), m_{\alpha, F}'(x) \right) = m_{\alpha, F}(x)$ because $m_{\alpha, F}(x)$ is irreducible. It follows that $m_{\alpha, F}(x) \mid m'_{\alpha, F}(x)$. We deduce that $m_{\alpha, F}'(x) = 0$. Otherwise, $\deg m_{\alpha, F}'(x) < \deg m_{\alpha, F}(x)$, so $m_{\alpha, F}(x)$ cannot divide $m_{\alpha, F}'(x)$. It follows from the definition of derivative of polynomials that$$m_{\alpha, F}'(x) = 0 \implies m_{\alpha, F}(x) = a_{m}x^{p^{m}} + a_{m-1}x^{p^{m-1}} + \cdots + a_{1}x^{p} + a_{0}.$$The condition $F = F^{p}$ implies that there exists $b_{0}, \dots, b_{m} \in F$ such that $a_{j} = b_{j}^{p}$. Then$$\begin{align*} m_{\alpha, F}(x) &= b_{m}^{p}x^{p^{m}} + b_{m-1}^{p}x^{p^{m-1}} + \cdots + b_{1}^{p}x^{p} + b_{0}^{p} \\ &= \left( b_{m}x^{p^{m-1}} + b_{m-1}x^{p^{m-1}} + \cdots + b_{1}x + b_{0} \right)^{p},\end{align*}$$and $m_{\alpha, F}(x)$ is not irreducible. We get a contradiction. Hence, $m_{\alpha, F}(x)$ is separable, and $\alpha$ is separable over $F$, so $K / F$ is separable and $F$ is perfect.