Let $X$ be a finite-dimensional normed space. Consider a non-empty convex set $C\subset X$ such that $0\notin C$. Notice that $C$ has a dense and countable subset $\{x_n\}$. $\forall n $ let
$C_n= \{x=\sum_{i=1}^{n}t_ix_i : t_i≥0, \sum_{i=1}^{n}t_i=1 \}$
It can be proved that $C_n$ is convex and compact and that $\bigcup_n C_n$ is dense in $C$.
How could be shown that for every $n$ there exists some $f_n \in X'$ such that $||f_n||=1$ and $f_n(x)≥0$ for all $x \in C_n$?
Well, the classic way to do this is to use the Hahn-Banach separation theorem, which works in infinite dimensional normed linear spaces, and in fact, locally convex topological vector spaces. But, in the case of finite-dimensions, you can prove it by alternate means.
Let $X = \mathbb{R}^n$ under the Euclidean norm, without loss of generality (by which I mean, there exists an isomorphism between $X$ under its own norm and this space, and conjugating the resulting functional on $\mathbb{R}^n$ by this isomorphism will yield a functional on $X$ that will do what you want).
Consider the (Euclidean) norm on $C$. It is a continuous function on a compact set, hence it attains a minimum at some point $c$. I claim the functional $f : x \mapsto c \cdot x$ is what we need: specifically a non-zero functional on $\mathbb{R}^n$ that sends every point in $C$ to a positive number. (We can then easily normalise $f$ afterwards.)
We may prove this fairly easily using the Variational Inequality. It says that, for any $d \in C$, we have, $$(0 - c) \cdot (d - c) \le 0 \implies c \cdot d \ge c \cdot c \ge 0 \implies f(d) \ge 0.$$ The functional $f$ is also non-zero, since $f(c) = c \cdot c = \|c\|^2 > 0$, since $c \neq 0$.