Non-dense subspace implies non-zero measure

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I am reading a proof of the Stone-Weierstrass theorem by De Branges, but I am having trouble understanding the following part.

Let $E$ be a locally compact Hausdorff space and $C_0(E, \mathbb{C})$ the continuous functions vanishing at infinity. For a vector subspace $S \subset C_0(E, \mathbb{C})$, define the set $M(S)$ of all signed measures $\mu$ on $E$ satisfying the following two properties: $\mu$ has a total variation of at most 1 and for all $f \in S$ we have $\int f \, d\mu = 0$. Equip $M(S)$ with the weak-*topology. Then if $S$ is not dense in $C_0(E, \mathbb{C})$, then $M(S)$ contains a non-zero measure.

I have tried the following. Since $S$ is not dense in $C_0(E, \mathbb{C})$, there exists a function that cannot be approximated by a net. I have tried constructing a measure from this function that satisfies the conditions of $M(S)$, but I did not manage to succeed. Also I tried, arguing by contraposition, showing that if $M(S)$ consists only of the zero measure, then $S$ would be dense in $C_0(E, \mathbb{C})$. For this I would have to show that any function in $C_0(E, \mathbb{C})$ is the limit of some net. However, I failed here also. Perhaps, I could use the Riesz-Markov-Kakutani theorem?

I have no clue on how to show this. Does anyone have a hint? Thank you very much :)