Let $E$ be a set of finite measure. I want to show that if $0 < p_1 < p_2 \leq \infty$, then $L^{p2} \subset L^{p1}$.
Approach: Split this into two cases ( when $p_2 < \infty$, and when $p_2 = \infty$ ). In the case when $p_2 < \infty$, I want to write $E = E_1 \cup E_2$ where $E_1 = \{ x \in E : |f(x)| \leq 1 \}$ and $E_2 =\{ x \in E : |f(x)| > 1 \}$. Using this, I want to express$\int_E |f|^p$ interms of $E_1$ and $E_2$ to prove the claim.
Any help on this would be appreciated.
If your space has measure 1 (probability space), you can use Jensen inequality for the case $p_2< \infty$ (the general case can be obtained from this). Consider the function $\phi (y)=|y|^{p_2/p_1}$ which is convex on $\mathbb{R}$. Now, you have $f\in L^{p_2}$ and you want to show that it is in $L^{p_1}$ as well (call E the space):
$$||f||_{p_1}^{p_2}=\phi(\int_{E}{|f|^{p_1}})\leq\int_{E}{\phi(|f|^{p_1}})=\int_{E}{|f|^{p_2}}=||f||_{p_2}^{p_2}.$$
While if $p_2=\infty$ you can just use that $\int_{E}|f|^{p_1}\leq \int_{E}||f||_{\infty}^{p_1}.$
Still, your approach to the problem is not wrong, to conclude you just need to observe that on $E_1$ the integral is bounded by $|E|$ while on $E_2$, because $f\geq 1$, you have no problem saying $|f|^{p_1}\leq |f|^{p_2}$ so you get $||f||_{p_1}\leq |E|+ ||f||_{p_2}$.
For the purpose of the exercise you can proceed in one or the other way, however it is interesting to see that the method I showed you implies an important consequence, that the inclusion of $L^{p_2}$ in $L^{p_1}$, when both are equipped with the topology given by the $L^p$ norm, is a continuous mapping: it is a linear and bounded operator.