An Orlicz norm is a norm

For triangle inequality, take $X,Y$ for which $\lVert\cdot\rVert_{\psi}$ is finite, and for a fixed $\varepsilon$, consider $u$ and $v$ such that $u-\lVert X\rVert_\psi<\varepsilon$, $v-\lVert Y\rVert_\psi<\varepsilon$, and $$\max\left\{E\left[\psi\left(\frac{|X|}u\right)\right],E\left[\psi\left(\frac{|Y|}v\right)\right]\right\}\leqslant 1.$$ Then \begin{align} \psi\left(\frac{|X+Y|}{u+v}\right)&\leqslant\psi\left(\frac{|X|+|Y|}{u+v}\right)&\phi\mbox{ is non-decreasing}\\ &=\psi\left(\frac u{u+v}\frac{|X|}u+\frac v{u+v}\frac{|Y|}v\right)&\\ &\leqslant \frac u{u+v}\psi\left(\frac{|X|}u\right)+\frac v{u+v}\psi\left(\frac{|Y|}v\right)&\mbox{ by convexity}. \end{align} This is what we do when we have the definition of the Orlicz norm.

Once can check that actually, the infimum is attained (taking a sequence $(u_n,n\geqslant 1)$ approaching the infinimum, and using Fatou's lemma). This will avoid the reasoning with $\varepsilon$.

Actually $\phi(x)$ can not be bounded, as long as it is not zero function. For convex functions, there is a super additive property if $\phi(0)\leq0$, which is $\phi(x)+\phi(y)\leq\phi(x+y)$, you can see Wiki page for this simple property. Now there exists $a\in R$ such that $\phi(a)>0$, then $\phi(na)>n\phi(a)$.