Number of injective field homomorphism Let $F$ and $L$ be two field extensions of $K$. The extension $F/K$ is finite of degree $n$ and the extension $L/K$ may be infinite also. 
I want to count the number of injective homomorphism from $F$ to $L$ where K is fixed. Is it true that the number of injective homomorphism from $F$ to $L$ where K is fixed  is less than or equal to $n$?
I know that the injective homomorphisms from $F$ to $L$ are lineraly independent over $L$. So, I'm trying to prove that if there are $n+1$ injective homomorphism, they must be linearly dependent. So,a contradiction. Is this approach correct?
 A: Yes it is true (and keeep in mind that a morphism of fields is automatically injective).  
The key point is that the set of $K$-linear  maps $f:F\to L$ is a vector space over the field $L$.
Indeed, for $\lambda \in L$ the map $\lambda f: F\to L \; $ is defined (you guessed it!) by $(\lambda f)(x)=\lambda \cdot (f(x))$ where the dot $\cdot$  is the product in the field $L$.  
Linear algebra then teaches us  that the dimension of that $L$-vector spase $\mathcal L_{K-lin }(F,L)$ is $n=dim_K(F)$  [see proof in Edit below]
The theorem of linear independence of homomorphisms then states that the set of $K$-algebra morphisms $Hom_{K-alg }(F,L)$ is a linearly independent subset $Hom_{K-alg }(F,L)\subset \mathcal L_{K-lin }(F,L)$ of the aforementioned $L$-vector space $\mathcal L_{K-lin }(F,L)$, so that of course $$\operatorname {card} Hom_{K-alg }(F,L)\leq n$$ just as you wished.  
Caveat
The set $Hom_{K-alg }(F,L)$ has no algebraic structure whatsoever: it is an unashamedly naked  set, very possibly empty .  
Edit
In order to answer user's question in the comments,  here is a proof that $\mathcal L_{K-lin }(F,L)$ has dimension $n$ over $L$:  
Choose a basis $a_1,...,a_n$ of $F$ over $K$.
Then the $K$-linear maps $f_i:F \to L$ defined by $f_i(\sum k_r a_r)= k_i$ are the required basis.
This is just a slight generalization of the usual concept of dual basis, which you recover if $L=K$ (which is a perfectly legitimate choice for $L$ in the question and in the answer!).   
Be very wary of the confusing fact that all $f_i$'s have values in $K$ but that there exist linear maps $f\in  \mathcal L_{K-lin }(F,L)$ capable of reaching any $\lambda\in L$: for example $(\lambda f_1)(a_1)=\lambda$ !
