I am working on a problem and I think I have an answer, but want to check it against others.
Let $p,q$ be as in title and define the norm on $L^p\cap L^q$ by $f\mapsto\|f\|_p+\|f\|_q$.
I have already shown $L^p\cap L^q\subset L^2$, that this intersection is dense, and that $L^p\cap L^q$ is complete.
Suppose there is a $C$ such that $$C(\|f\|_p+\|f\|_q)\leq \|f\|_2.$$
Then its clear that $L^2=L^p\cap L^q$, right? I mean $f\in L^2$ will clearly be in $L^p\cap L^q$ right? Or is it that when we are saying $L^2=L^p\cap L^q$ we are saying that as two topological spaces they are the same.
Finally, I need to find an example of this. If I let $X=[0,1]$ and consider Lesbegue measure then $L^p=L^r$ for every $1< r$ and so it is clear that $L^p\cap L^q=L^2$.
Let $f \in L^{2}$. there exists a sequence $(f_n)$ in $L^{p}\cap L^{q}$ such that $\|f_n -f\|_2 \to 0$. By the given inequality we see that $\|f_n- f\|_p+\|f_n -f\|_q \to 0$. Hence $(f_n)$ is Cauchy in $L^{p}$ as well as $L^{q}$. Let $f_n \to g$ in $L^{p}$ and $f_n \to h$ in $L^{q}$. Convergence in any $L^{p}$ implies almost everywhere convergence of a subsequence. From this it follow that $f=g=h$ a.e.. This proves that $f \in L^{p}\cap L^{q}$.
For Lebesgue measure on $[0,1]$ no two $L^{p}$ spaces are the same, so your example fails. Take a degenerate measure on $[0,1]$, say $\mu =\delta_0$. Then all $L^{p}$ spaces are equal since any real valued function belongs in any of these spaces. Also $\|f||_p=|f(0)|$ so above inequality holds with $C=\frac 1 2$.