Show that in a Polish space $E$ the Dirac measures are the only non-zero Borel measures $\mu$ which take only the values $0$ and $1$. Show that in a Polish space $E$ the Dirac measures are the only non-zero Borel measures $\mu$ which take only the values $0$ and $1$.
This is an exercise from Bauer's Measure and Integration Theory. There is a hint to the problem. Hint: Show that the system of all compact $K \subset E$ such that $\mu(K)=1$ is $\cap$-stable and investigate the intersection of all its sets. 
I have shown that the system is $\cap$-stable, and I know from the finite intersection property that the intersection of all its sets is nonempty, but I don't know how to progress from here. I would greatly appreciate any help.
 A: The key property here is that, on a Polish space, any finite Borel measure $\mu$ is inner regular. This means that for any open set $V$ and any $\varepsilon>0$, there is a compact set $K \subset V$ with $\mu(V) - \mu(K) < \varepsilon$. This is Lemma 26.2 of Bauer. To emphasize the importance of this property I will show the following more general claim:

Proposition: Let $E$ be a Hausdorff topological space. Then the only non-zero inner regular Borel measures that take only the values $0$ and $1$ are the Dirac measures.

Proof:
Let $\mu$ be such a measure on $E$. Following the hint, let $\mathcal{C}$ be the collection of compact sets of measure $1$. Since $\mu$ is non-zero we have $\mu(E) = 1$. By the inner regularity of $\mu$, there exists a compact set $K$ with $\mu(K)=1$, so $\mathcal{C}$ is non-empty.
We show that $\mathcal{C}$ has the finite intersection property: the intersection of any finite subcollection of $\mathcal{C}$ is non-empty. In fact, we will prove that the intersection of any finite subcollection of $\mathcal{C}$ has measure $1$. We procede by induction on the number of elements of the subcollection. If it consists of only one set, the claim is trivial. Let $\{ K_1, \ldots, K_n \}$ be a subcollection of $\mathcal{C}$, and assume that the claim is true for a subcollection of $n-1$ elements. Then 
$$
\mu( \cap_{i=1}^n K_i ) = \mu( K_1 \cap ( \cap_{i=2}^n K_i)) = 
\mu(K_1) + \mu(\cap_{i=2}^n K_i) - \mu( K_1 \cup (\cap_{i=2}^n K_i)) 
= 1 + 1 - 1 = 1,
$$
so $\cap_{i=1}^n K_i$ has measure $1$ and, in particular, is non-empty.
Since $\mathcal{C}$ is a family of compact sets with the finite intersection property, its intersection $I$ is non-empty. Let $x$ be any point in $I$. We show that $\mu ( \{ x \} ) = 1$ (and this means that $\mu$ is the Dirac measure at $x$). Assume that this is not the case. Then $\mu( \{ x \}) =0$, so $\mu( E \setminus \{ x \} ) =1$. Since $E \setminus \{ x \}$ is open, there is a compact set $K \subset E \setminus \{ x \}$ with $\mu(K) =1$. But then by definition of $I$ we have $I \subset K  \subset E \setminus \{ x \}$, which is a contradiction since $x \in I$.
