How Hahn-Banach theorem implies that the dual space is non-trivial? Why does the theorem of Hahn-Banach implies that the dual space is not empty ($X^*\neq\emptyset)$ ?
Is there an important corollary which I've missed ? 

If $X \neq \{0\}$(!), you have some $x \in X$ with $x \neq 0$ which implies that $\|x\| \neq 0$ by definiteness of the norm. 
You can define now the functional $\varphi(\alpha x) = \alpha\|x\|$ on $\operatorname{span} x$. Now,  $\|\alpha x\| = 1$ iff $|\alpha| = \frac{1}{\|x\|}$. This implies that 
$$
|\varphi(\alpha x)| = |\alpha \varphi(x)| = 1
$$
for all $\alpha x$ with norm 1.
In particular $\varphi$ is continuous with operator norm $\|\varphi\| = 1$ and by Hahn-Banach there exists some extension $\tilde \varphi \in X^*$ with $\tilde \varphi \neq 0$, since $\tilde \varphi(x) = \varphi(x) = \|x\| \neq 0$. Alternatively, you see that $\tilde \varphi \neq 0$ since $\|\tilde \varphi\| = \|\varphi\| = 1 \neq 0$, once again by definiteness of the (operator) norm.
