= Solution
The interpolated exponent is strictly between $p_0$ and $p_1$. Apply <Hölder's inequality> to $|f|^p=|f|^{p(1-\theta)}|f|^{p\theta}$ with <conjugate exponents> $p_0/[p(1-\theta)]$ and $p_1/(p\theta)$. Their reciprocals sum to one, so
$$
\boxed{\|f\|_p\leq\|f\|_{p_0}^{1-\theta}\|f\|_{p_1}^{\theta}.}
$$
This gives $L^{p_0}\cap L^{p_1}\subseteq L^p$.
For the other inclusion take any $a>0$ and use the <threshold decomposition between Lp spaces>
$$
g=f\mathbf1_{\{|f|>a\}},\qquad
h=f\mathbf1_{\{|f|\leq a\}}.
$$
Because $p_0-p<0<p_1-p$,
$$
\int|g|^{p_0}\leq a^{p_0-p}\int|f|^p,\qquad
\int|h|^{p_1}\leq a^{p_1-p}\int|f|^p.
$$
Thus $g\in L^{p_0}$, $h\in L^{p_1}$ and $f=g+h$, proving $L^p\subseteq L^{p_0}+L^{p_1}$. Raising the two <norm> bounds to powers $1-\theta$ and $\theta$ gives a power of $a$ equal to
$$
(1-\theta)(1-p/p_0)+\theta(1-p/p_1)=0,
$$
and a power of $\|f\|_p$ equal to one. Hence
$$
\boxed{\|g\|_{p_0}^{1-\theta}\|h\|_{p_1}^{\theta}\leq\|f\|_p.}
$$
For later use, choose $a=\|f\|_p$ when $f\ne0$. Each endpoint <norm> is then at most $\|f\|_p$, so $\|f\|_{L^{p_0}+L^{p_1}}\leq2\|f\|_p$ for the <sum of Lp spaces> <norm>. If $f=0$, take $g=h=0$.
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