For any measurable $f\ne0$, and for any $\varepsilon$, there exists a measurable subset $A$ with positive measure (and so $\mu(A)\geq1$) such that $|f|\geq\|f\|_\infty-\varepsilon$ on $A$. Then $ \|f\|_q=\left(\int|f|^q\right)^{1/q}\geq\left(\int_A|f|^q\right)^{1/q} \geq\left(\int_A(|f|_\infty-\varepsilon)^q\right)^{1/q}=(\|f\|_\infty-\varepsilon)\,\mu(A)\geq\|f\|_\infty-\varepsilon. $ As $\varepsilon$ was arbitrary, the first inequality is proven.
For the second inequality, assume first that $\|f\|_q=1$. This implies, by the first inequality, that $|f|\leq1$ a.e. Then, using $p\leq q$ (and so $|f|^p\geq|f|^q$), $ \|f\|_p=\left(\int|f|^p\right)^{1/p}\geq\left(\int|f|^q\right)^{1/p}=1^{1/p}=1=\|f\|_q $ The case $\|f\|_q\ne1$ follows easily.
The third inequality is a particular case of the second one.