field extension problem I'm suppossed to use an example to show the following statement.
If F over K is galois but not algebraic and L is an intermediate field between K and F, then F over L is not galois. 
Any help at all would be greatly appreciated.
 A: First let me say that I am interpreting your question according to the definition of Galois for transcendental extensions given by Adrian Barquero (this is not my preferred definition: see below).  Second, interpreting it that way it seems that you have a quantifier error: you ask for a Galois transcendental extension $F/K$ such that for every subextension $L$ of $F/K$, $F/L$ is not Galois.  But this can't be: take for instance $L = F$ or more generally let $G$ be any finite subgroup of $\operatorname{Aut}(F/K)$ and let $L$ be the fixed field $F^G$.  Then $F/F^G$ is a finite Galois extension.  If $F/K$ is transcendental and Galois then $\operatorname{Aut}(F/K)$ must be infinite and thus there are infinitely many examples like this.
I will assume you meant to ask: find some Galois transcendental extension $F/K$ and *some subextension $L$ such that $F/L$ is not Galois.
For this: $K = \mathbb{Q}$ and $F = \mathbb{Q}(t)$.  Then the only elements of $F$ fixed by every element of $\operatorname{Aut}(K/\mathbb{Q}) = \operatorname{Aut}(K) = PGL_2(\mathbb{Q})$ acting by linear fractional transformations are the elements of $\mathbb{Q}$.  So $F/K$ is Galois according to the definition Adrian provides above.
However, consider the intermediate field $L = \mathbb{Q}(t^3)$.  Then $F/L$ is not Galois because $\mathbb{Q}$ does not contain a primitive $3$rd root of unity.
You can create similar examples with ground field $K = \mathbb{C}$ by choosing a finite subgroup $G$ of $PGL_2(\mathbb{C}) = \operatorname{Aut}(\mathbb{C}(t)/\mathbb{C})$ which contains a non-normal subgroup $H$.  Then take $F = \mathbb{C}(t)^H$ and $L = \mathbb{C}(t)$.  For instance, you can take $G$ to the be icosahedral group $A_5$, which is nonabelian simple, so has plenty of non-normal subgroups.
Note that in this rough note I attempt to study transcendental Galois extensions.  The property of an extension $F/K$ that $F^{\operatorname{Aut}(F/K)} = K$ I call weakly Galois.  My definition of a Galois extension is precisely that the examples you seek do not occur: namely for every intermediate extension $L$, also
$F^{\operatorname{Aut}(F/L)} = L$.  So in my terminology, for any infinite field $K$,
the rational function field $K(t)$ is weakly Galois but not Galois.
In fact I have made a conjecture that a transcendental extension $F/K$ is Galois iff $F$ is algebraically closed of characteristic $0$.  One implication is easy; the other one has been an open question on MathOverflow for more than a year.
A: If you use the definition of Galois extension that I quoted in the comments then you can look at the field extension $K(x) / K$ where $K$ is an infinite field. In this case this field extension is Galois but not algebraic over $K$. So my suggestion is for you to look at exercises $6$ and $9$ in section $2$, chapter $5$ of Thomas Hungerford's algebra book.
In exercise $6$ you are guided in a series of steps to prove some important facts about this field extension and about the automorphism group. And exercise $9$ guides you to prove that this extension is Galois.
