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Combine this with Autoleech's proof that there is no counterexample for locally compact spaces, and we will have a proof that there are no measurable cardinals - finally!
To maybe clarify Andrej Bauer's comment, the 2nd part of the displayed formula is equivalent to the form $\forall x,y,n (\phi\wedge R^n\rightarrow\ldots)$.
@Gerald Edgar: good question, maybe argue that we can make a real number belong to each $V_n$ by specifying more and more of its binary expansion. We can even take breaks, i.e., after we have made $r$ start as $0.r_0\cdots r_{k_1}$ and thereby ensured $r\in V_1$ we can append $r_{k_1+1}\cdots r_{\ell}$ so as to make sure $r\ne p_1$. Then we add $r_{\ell+1}\cdots r_{k_2}$ to make $r\in V_2$ and so on. This kind of thing also shows $\cap_n V_n$ is size continuum.
