Let $\Phi : [0, \infty) \rightarrow [0, \infty)$ a convex, strictly incrasing function, with $\Phi(0)=0$.
Let $L^{\Phi}(0,1)=\{f:(0,1)\rightarrow\mathbb{R} \text{ measurable}:\int_0^1\Phi\left( \frac{|f(t)|}{\lambda} \right) \, dt< \infty \text{ for some } \lambda>0\}$
Let $\lVert f \rVert_{\Phi} = \inf \{\lambda > 0 : \int_0^1\Phi\left( \frac{|f(t)|}{\lambda} \right)\, dt < 1\}$
Show that $\lVert f \rVert_{\Phi}<\infty, \forall f \in L^{\Phi}(0,1)$.
My try: $f$ must have finite integral, for if it is not the case, since $\Phi$ must be eventually always be larger than $1$ for every $x>M$ for some $M$ (easy to prove), then we have
$$\int_0^1\Phi\left( \frac{|f(t)|}{\lambda} \right) \, dt \geq \int_{E_1}\frac{|f(t)|}{\lambda} = \infty, \forall \lambda>0$$ where $$E_1=\{x\in(0,1):|f(x)|\geq \lambda M\}$$ so $f$ wouldn't belong to $L^{\Phi}(0,1)$ in the first place. On the other hand, is $f$ is integrable, then, letting $\lambda \rightarrow \infty$, the argument of $\Phi$ tends to zero, and $\Phi$ must be close to $0$ as its argument approaches $0$. Therefore the whole integral tends to zero.
More specifically, we can suppose $\exists \, \delta: \Phi(x)<\epsilon\ \forall x<\delta $, and that $\frac{|f(x)|}{\lambda} < \delta$ everywhere for $\lambda$ large enough, and so
$$\int_0^1\Phi\left( \frac{|f(t)|}{\lambda} \right) \, dt \leq \int_0^1 \delta\, dt$$ for $\lambda$ large enough.
Is it correct? My doubt is, mainly, that it feels awkward to bound those integrals like that without assuming continuity on $\Phi$ or some stronger condition that is usually assumed.
Your claim that $|f (x)|/\lambda <\delta $ everywhere is not correct. However your general idea is correct. To conclude, by the monotone convergence theorem, it suffices that pointwise the function converges to zero, and it does so monotonously.