Simple proof of Hahn-Banach in finite dimensional space Let $E$ be a finite dimensional Banach space and $C \subseteq E$ a non empty and convex set such that $0 \notin C$. I have to use the following steps to prove that $\exists T \in E'$ such that $T(x) \geq 0 \, \forall x \in C$:
1) Let $(x_n)_{n=1}^\infty\subset C$ dense and consider $C_n = \text{co}(x_1,\dots,x_n)$. Prove that $C_n$ is compact and $D=\bigcup_n C_n$ is dense in $C$.
2) Prove that exists $T_n \in E'$ such that $||T_n||=1$ and $T_n(x) \geq 0$ for all $x \in C_n$.
3) Prove that exists $T \in E'$ such that $||T||=1$ and $T(x) \geq 0$ for all $x \in C$.

My solution:
1) Let $f\colon \mathbb{R}^n \to E$, $f(\lambda_1,\dots,\lambda_n)=\sum_{i=1}^n \lambda_ix_i$ then $C_n = f(S)$ with $S = \{\lambda \in \mathbb{R}^n \colon \lambda \geq 0, \, \sum_{i=1}^n \lambda_i = 1\}$. hence $C_n$ is compact. $D=\bigcup_n C_n$ is dense because $(x_n) \subset D$.
2) I don't know how to do this.
3) $E'$ is finite dimensional and $T_n$ lives on the closed ball, then exists a subsequence $(T_{n_k})_k$ which converges to an operator $T$ with $||T||=1$. Now if $x \in D$ then $\exists N>0$ such that $x \in C_n$ for all $n \geq N$, then we have $$T_{n_k}(x) \geq 0 $$
for $k$ big. Taking limit we get $T(x) \geq 0$ for all $x \in D$ and this extend to all $C$ because $D$ is dense on $C$.
I need help with the second part.
any help will be appreciated.

Set $C_n$ is compact and set $\{0\}$ is closed. By Hahn-Banach separation theorem, we have some $\tilde{T}_n$ such that $\tilde{T}_n(x)\geq 0= \tilde{T}_n(0)$. Let $T_n=\tilde{T}/\|\tilde{T}\|$.
