Same proof, less technical. Suppose $\mu$ is a measure taking exactly values $0,1$. We claim $\mu$ is a dirac measure.
Let $U$ be the union of all open sets of measure zero. Since $\mathbb C$ is 2nd countable, $U$ is the union of coutably many open sets of measure zero, so $\mu(U) = 0$. Let $F$ be the complement of $U$. Then $F$ has measure $1$. I claim that $F$ is a singleton. If not, there are two points $x \
e y$ in $F$, and thus there are disjoint open neighborhoods $U_x, U_y$ of them. Since $U_x$ is not a subset of $U$, $\mu(U_x) = 1$. Similarly $\mu(U_y)=1$. So in fact $\mu(\mathbb C) \ge 2$ a contradiction. Thus $F$ is a singleton, $\mu(F)=1$, $\mu(\mathbb C \setminus F) = 0$.
So I guess the point of doing another proof is: you don't need the exact structure of $\mathbb C$, merely that it is a 2nd countable Hausdorff space.