$\implies$: Given $\alpha \in L$, it is algebraic over $K$ since $L / K$ is algebraic. So there exist $a_{0}, \dots, a_{n-1} \in K$ such that $$ a(\alpha) = \alpha^{n} + a_{n-1} \alpha^{n-1} + \cdots + a_{0} = 0. $$ Hence $\alpha$ is algebraic over $F(a_{0}, \dots, a_{n-1})$, and $$ [F(\alpha, a_{0}, \dots, a_{n-1}) : F(a_{0}, \dots, a_{n-1})] < \infty. $$ Since $K / F$ is algebraic, each $a_{i}$ is algebraic over $F$, so $[F(a_{i}) : F] < \infty$ for all $i$. Thus $$ [F(a_{0}, \dots, a_{n-1}) : F] < \infty. $$ By Theorem 1207, we write $$ [F(\alpha, a_{0}, \dots, a_{n-1}) : F] = [F(\alpha, a_{0}, \dots, a_{n-1}) : F(a_{0}, \dots, a_{n-1})][F(a_{0}, \dots, a_{n-1}) : F], $$ then we can conclude that $[F(\alpha, a_{0}, \dots, a_{n-1}) : F] < \infty$. Since $F \subseteq F(\alpha) \subseteq F(\alpha, a_{0}, \dots, a_{n-1})$, it follows that $[F(\alpha) : F] < \infty$. Then by Proposition 1403, $\alpha$ is algebraic over $F$. Since $\alpha$ is arbitrary, $L / F$ is algebraic.

$\impliedby$: Since $L / F$ is algebraic and $K \subseteq L$, $K / F$ is also algebraic. Moreover, every $\alpha \in L$, there exist $a_{0}, \dots, a_{n-1} \in F$ such that $$ a(\alpha) = \alpha^{n} + a_{n-1}\alpha^{n-1} + \cdots + a_{0} = 0. $$ Since $F \subseteq K$, each $a_{i} \in K$, hence $a(x) \in K[x]$, so $\alpha$ is algebraic over $K$. Therefore, $L / K$ is algebraic.