This question has been asked here, but the (upvoted) answer doesn't seem to be correct (see my comment below). Hence I re-ask the question:
If $X$ is a compact Hausdorff space, then is its topology precisely the weak topology induced by $C(X)$ (the set of continuous functions $X\to\mathbb K$ where $\mathbb K$ is $\mathbb R$ or $\mathbb C$)?
Let's set the notation: Keep denoting the original space by $X$, whereas use $X_w$ for the weak topology. It's easy to see that $X$ is finer than $X_w$. Conversely, let $U$ be open in $X$. Now by Urysohn ($X$ is normal), we have a continuous function $f\colon X\to [0, 1]$ such that $f$ is $0$ on $X\setminus U$. If I had that $f$ is $0$ precisely on $X\setminus U$, then I'd be done, for then $U$ would be the inverse image of closed $\{0\}$ under $\iota\circ f\colon X\to\mathbb C$ ($\iota$ is the inclusion $[0, 1]\hookrightarrow\mathbb C$) which being continuous means that it's also continuous on $X_w\to\mathbb C$.
But $f^{-1}(\{0\})$ can be larger than $X\setminus U$, right? How to rectify the situation?
I talked with my advisor, and he sorted this out:
Let $U$ be open in $X$. W.l.o.g., let $U\subsetneq X$. Let $x\in X$. It suffices to produce a set open $V_w$ in $X_w$, such that $x\in V_w\subseteq U$. By Urysohn (since $X$ is normal), there exists a continuous function $f\colon X\to\mathbb K$ such that $f(x) = 0$ and $f(X\setminus U) = \{1\}$. Now, $V := f^{-1}(B_{1/2}(0))$ does the job since $f\colon X_w\to \mathbb K$ is continuous.