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 ? 
 A: The dual space is always non-empty, as it contains the zero functional. The Hahn-Banach theorem implies that if $X \neq \{0\}$, then also $X^* \neq \{0\}$. 
Choose a non-zero vector $a \in X$. Denote the subspace $Y := \mathrm{span} (a) \subseteq X$ and the bounded functional $\varphi \in Y^*$ defined by $\varphi(a) = 1$. By Hahn-Banach, it can be extended to some bounded functional on the whole space $\overline{\varphi} \in X^*$, which is non-zero (since $\overline{\varphi}(a) \neq 0$), hence $X^* \neq \{0\}$. 
A: 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.
A: Let $X$ be a separated locally convex space. If $x\not=0$, then we can separate the compact $\{0\}$ and the closed $\{x\}$, i.e., there exists $f\in X^{*}$ such that $f(0)<f(x)$. Since $f(0)=0$, we have $f(x)>0$. In particular, $f\not=0$. So, $X^{*}\not=\{0\}$.
