Theorem 3.4. Normal extensions remain normal under lifting. If $K \supset E \supset k$ and $K$ is normal over $k$, then $K$ is normal over $E$. If $K_1$, $K_2$ are normal over $k$ and are contained in some field $L$, then $K_1 K_2$ is normal over $k$, and so is $K_1 \cap K_2$.
Proof. For our first assertion, let $K$ be normal over $k$, let $F$ be any extension of $k$, and assume $K$, $F$ are contained in some bigger field. Let $\sigma$ be an embedding of $KF$ over $F$ (in $F^*$). Then $\sigma$ induces the identity on $F$, hence on $k$, and by hypothesis its restriction to $K$ maps $K$ into itself. We get $(KF)^\sigma = K^\sigma F^\sigma = KF$ whence $KF$ is normal over $F$.
Lang proves the above theorem in his book Algebra. The only first part of the proof seems to be weird, namely that "lifting respects normal extensions".
Indeed, he has shown that any embedding $\sigma$ of $KF$ in $F^a$ over $F$ gives an automorphism. And this is one of the equivalent definitions of normal extensions.
But we also have to show that $KF$ is algebraic over $F,$ which I do not see in his proof (because the definition of normal extension requires an algebraic extension with certain properties).
How to show that $KF$ is algebraic over $F$?
Would be very grateful for any help!
Pulling down the comment by @ArturoMagidin.
Lang has already shown that you can lift algebraic extensions: if $K/k$ is algebraic, and $F$ is any extension of $k$, then $KF/F$ is algebraic.
For reference, see Proposition 1.7 of Chapter V (page 228, third edition):