$\implies$ : 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.