Varieties over a field $K$ are also varieties over any subfield of $K$. Suppose that $f:X\longrightarrow\text{Spec} K$ is a variety over $K$, namely $X$ is an integral, separated $K$-scheme of finite type. Now if $L$ is a subfield of $K$, it is clear that there exists a morphism of affine schemes $g:\text{Spec K}\longrightarrow\text{Spec L}$.
At this point one can consider the morphism $g\circ f: X\longrightarrow\text{Spec }   L$ to conlude that $X$ is also a variety over $L$. By this reasoning, it seems that any variety over $K$ can be viewed as a variety oover any subfield, but this sentence is clearly false. Where is the mistake?

A variety is not just the scheme $X$ but also the structural morphism $f:X\rightarrow Spec(K)$. To say that this variety can be defined over a subfield $L$ of $K$, as I understand, it means that there exists a variety $Y$ defined over $L$ such that $X=Y\times_{Spec(L)} Spec(K)$ and $f$ is the projection over $Spec(K)$. 
Now, note that if you are considering $X$ as a variety over $L$ via $Spec(K)\rightarrow Spec(L)$ then it is not true in general that $X$ (as variety over $K$) is the fibered product $X\times_{Spec(L)}Spec(K)$.
