Let $f$ be a function in both $L^{p_1}(0,\infty)$ and $L^{p_2}(0,\infty)$ with $0 < p_1 < p_2 < \infty$. I want to show that for all $p$ with $p_1 < p < p_2$, we have that $f \in L^p(0,\infty)$.
I'm aware of the result that when $\mu(X)$ is finite we have that $p_1 < p_2$ implies $L^{p_2} \subset L^{p_1}$. Since it's infinite in this case, I can't use this method.
Write $f = f_1 +f_2$ with $f_1 = f 1_{\{|f|\le 1\}}$ and $f_2 = f 1_{\{|f|\le 1\}}$. Note that $\|f\|_{L^p} \le \|f_1\|_{L^p}+ \|f_2\|_{L^p}$ by the $\Delta$-inequality. Thus it remains to show that $f_1,f_2 \in L^p$. For this use that $|f_1|^p \le |f|^{p_1}$ and $|f_2|^p \le |f|^{p_2}$ in order to conclude that, in fact, $f_1,f_2 \in L^p$.
Alternatively, we can use the Hölder inequality: Take $$r = (p_2-p_1)/(p-p_1)$$ and note that $r>1$ and $p = p_1/r' + p_2/r$, where $r'$ is the conjugate exponent, i.e. $1/r+1/r' =1$. By applying the Hölder inequality we get $$\int |f|^p \, d \mu = \int |f|^{p_1/r'} \cdot |f|^{p_2/r} \, d \mu \le \left( \int |f|^{p_1} \, d \mu \right)^{1/r'} \left( \int |f|^{p_2} \, d \mu \right)^{1/r}.$$ That is $$\|f\|_{L^p} \le \|f\|_{L^{p_1}}^{\frac{p_2-p}{p_2/p_1-1}} \|f\|_{L^{p_2}}^{\frac{p-p_1}{1-p_1/p_2}}.$$