if $k$ is any field then it is algebraically closed in $k(x)$. What if we take a nontrivial extension of $k$, i.e. $k'/k$, is $k$ still algebraically closed in $k'(x)$, maybe or never ? ($x$ is not algebraic over $k$)
Because there is an argument in a proof if $[E:k]=\infty$ then consider the infinite chain $k\subset E_1\subset E_2\subset\dots$ (distinct subextensions of $E)$ then there is an induced chain $k(x)\subset E_1(x)\subset E_2(x)\subset\dots$. the author argues in the following way:
$E_i$ is algebraically closed in $E_i(x)$ but not in $E_{i+1}(x)$ so new chain also consists of distinct elements.
the purpose of the proof is to show that $[E(x):k(x)]=[E:k]$
If $k$ is not algebraically closed in $k'$ (for instance, in the extreme case where $k'/k$ is an algebraic extension) then clearly $k$ is not algebraically closed in $k'(x)$: any $a\in k'$ which is algebraic over $x$ is also an element of $k'(x)$ which is algebraic over $k$.