The problem is as follows.
My solution : Let $t \geq 0$ be arbitrary. Then
$E(e^{tX})=\int_{-\infty}^{\infty} e^{tx}dF(x) \geq \int_{0}^{\infty} e^{tx}dF(x) \geq \int_{0}^{\infty} dF(x)=P\{X \geq 0\}$
Hence, $P\{X \geq 0\} \leq \inf \{E(e^{tX}); t \geq 0\} \leq E(e^{0 \cdot X})=1$
I'd appreciate if anyone checks it and tell me whether there's any technicalities that I may have missed. Do I need to show that $P\{X \geq 0\} \leq \inf \{E(e^{tX}); t \geq 0\}$ explicitly (by contradiction, for example)? Or is it fine to imply it directly?
The proof seems fine.
You have proven that $\forall t \ge 0, E(e^{tX}) \ge P\{X \ge 0\}$, which implies $$\inf\{ E(e^{tX}):t \ge 0\} \ge P\{ X \ge 0\}$$