Let $(X,\tau)$ be a compact topological space, and let $\large\tau'$ be the weak topology generated by some subset $S \subset \mathcal{C}(X,\mathbb{R})$ which seperates points.
I need to show that $\large\tau$ = $\large\tau'$.
I tried to show first that with respect to $\large\tau'$ it follows that $S=\mathcal{C}(X,\mathbb{R})$, i.e $f:X\rightarrow R $ is continuous with respect to $\large\tau'$ iff $f \in S$ .
But I'm not sure if this statement is correct (if it's correct I know how to finish the proof using it)
In some texts, compact means compact and Hausdorff. I will assume that $X$ is compact and Hausdorff. Then $(X,\tau')$ is also Hausdorff: I will let you check this, using the fact that $S$ separates points.
By definition $\tau'\subseteq \tau$. Consider the identity map $i: (X,\tau)\to (X, \tau')$. This is a continuous bijection. It maps closed sets to closed sets because closed subsets are compact (and continuous images of compact sets are compact, hence also closed). Hence, its inverse is continuous. This says that $\tau\subseteq \tau'$.