# A question about separable extension

We say $K$ is separably generated over $k$ if there exists a transcendence basis $\{x_i,i\in I\}$ of $K/k$ such that $K/k(x_i,i\in I)$ is a separable algebraic extension.

We say $K$ is separable over $k$ if every subextension $K'/k\subset K/K$ with $K'$ finitely generated over $k$, the extension $K'/k$ is separably generated.

My quesition is why not define $K/k$ be separable if every subextension (not necessarily finitely generated) is separably generated?

Defining separability the way you propose implies that separability and being separably generated are the same thing. Because $K/k$ is a subexetension of $K/k$. (Of course you could require, that one only considers proper subextensions -- but on what basis?)
However here is an informative example: let $x$ be transcendental over $k$ and consider $K=k(x,x^{1/p},x^{1/p^2},\ldots )$, where $k$ has characteristic $p>0$. Then $K/k$ is not separably generated, because the transcendence degree of the extension is $1$: being separably generated would therefore imply that $K=k(t)$ for some $t$, which is impossible.
On the other side $K$ is the union of a chain of rational function fields -- namely $k(x^{1/p^n})$ -- all of which are separable. So one "expects" $K$ also to be separable.