Write $K$ as a directed colimit $K = \colim_i K_i$ of finitely generated field extensions $k \subset K_i$. By definition $K$ is separable if and only if each $K_i$ is separable over $k$, and by Lemma \ref{lemma-colimit-differentials} we see that $K \otimes_k \Omega_{k/\mathbf{F}_p} \to \Omega_{K/\mathbf{F}_p}$ is injective if and only if each $K_i \otimes_k \Omega_{k/\mathbf{F}_p} \to \Omega_{K_i/\mathbf{F}_p}$ is injective. Hence we may assume that $K/k$ is a finitely generated field extension. \medskip\noindent Assume $k \subset K$ is a finitely generated field extension which is separable. Choose $x_1, \ldots, x_{r + 1} \in K$ as in Lemma \ref{lemma-generating-finitely-generated-separable-field-extensions}. In this case there exists an irreducible polynomial $G(X_1, \ldots, X_{r + 1}) \in k[X_1, \ldots, X_{r + 1}]$ such that $G(x_1, \ldots, x_{r + 1}) = 0$ and such that $\partial G/\partial X_{r + 1}$ is not identically zero. Moreover $K$ is the field of fractions of the domain. $S = K[X_1, \ldots, X_{r + 1}]/(G)$. Write $$ G = \sum a_I X^I, \quad X^I = X_1^{i_1}\ldots X_{r + 1}^{i_{r + 1}}. $$ Using the presentation of $S$ above we see that $$ \Omega_{S/\mathbf{F}_p} = \frac{ S \otimes_k \Omega_k \oplus \bigoplus\nolimits_{i = 1, \ldots, r + 1} S\text{d}X_i }{ \langle \sum X^I \text{d}a_I + \sum \partial G/\partial X_i \text{d}X_i \rangle } $$ Since $\Omega_{K/\mathbf{F}_p}$ is the localization of the $S$-module $\Omega_{S/\mathbf{F}_p}$ (see Lemma \ref{lemma-differentials-localize}) we conclude that $$ \Omega_{K/\mathbf{F}_p} = \frac{ K \otimes_k \Omega_k \oplus \bigoplus\nolimits_{i = 1, \ldots, r + 1} K\text{d}X_i }{ \langle \sum X^I \text{d}a_I + \sum \partial G/\partial X_i \text{d}X_i \rangle } $$ Now, since the polynomial $\partial G/\partial X_{r + 1}$ is not identically zero we conclude that the map $K \otimes_k \Omega_{k/\mathbf{F}_p} \to \Omega_{S/\mathbf{F}_p}$ is injective as desired. \medskip\noindent Assume $k \subset K$ is a finitely generated field extension and that $K \otimes_k \Omega_{k/\mathbf{F}_p} \to \Omega_{K/\mathbf{F}_p}$ is injective. (This part of the proof is the same as the argument proving Lemma \ref{lemma-characterize-separable-field-extensions}.) Let $x_1, \ldots, x_r$ be a transcendence basis of $K$ over $k$ such that the degree of inseparability of the finite extension $k(x_1, \ldots, x_r) \subset K$ is minimal. If $K$ is separable over $k(x_1, \ldots, x_r)$ then we win. Assume this is not the case to get a contradiction. Then there exists an element $\alpha \in K$ which is not separable over $k(x_1, \ldots, x_r)$. Let $P(T) \in k(x_1, \ldots, x_r)[T]$ be its minimal polynomial. Because $\alpha$ is not separable actually $P$ is a polynomial in $T^p$. Clear denominators to get an irreducible polynomial $$ G(X_1, \ldots, X_r, T) = \sum a_{I, i} X^I T^i \in k[X_1, \ldots, X_r, T] $$ such that $G(x_1, \ldots, x_r, \alpha) = 0$ in $L$. Note that this means $k[X_1, \ldots, X_r, T]/(G) \subset L$. We may assume that for some pair $(I_0, i_0)$ the coefficient $a_{I_0, i_0} = 1$. We claim that $\text{d}G/\text{d}X_i$ is not identically zero for at least one $i$. Namely, if this is not the case, then $G$ is actually a polynomial in $X_1^p, \ldots, X_r^p, T^p$. Then this means that $$ \sum\nolimits_{(I, i) \not = (I_0, i_0)} x^I\alpha^i \text{d}a_{I, i} $$ is zero in $\Omega_{K/\mathbf{F}_p}$. Note that there is no $k$-linear relation among the elements $$ \{x^I\alpha^i \mid a_{I, i} \not = 0 \text{ and } (I, i) \not = (I_0, i_0)\} $$ of $K$. Hence the assumption that $K \otimes_k \Omega_{k/\mathbf{F}_p} \to \Omega_{K/\mathbf{F}_p}$ is injective this implies that $\text{d}a_{I, i} = 0$ in $\Omega_{k/\mathbf{F}_p}$ for all $(I, i)$. By Lemma \ref{lemma-derivative-zero-pth-power} we see that each $a_{I, i}$ is a $p$th power, which implies that $G$ is a $p$th power contradicting the irreducibility of $G$. Thus, after renumbering, we may assume that $\text{d}G/\text{d}X_1$ is not zero. Then we see that $x_1$ is separably algebraic over $k(x_2, \ldots, x_r, \alpha)$, and that $x_2, \ldots, x_r, \alpha$ is a transcendence basis of $L$ over $k$. This means that the degree of inseparability of the finite extension $k(x_2, \ldots, x_r, \alpha) \subset L$ is less than the degree of inseparability of the finite extension $k(x_1, \ldots, x_r) \subset L$, which is a contradiction.