I'm currently reading Rogers and Williams book on diffusions, Markov processes, and martingale. Using ordinal techniques, they are able to prove that if $X$ is a càdlàg process with values in a separable metric space and $K$ is a compact set, then the hitting time of $X$ in $K$ is a stopping time, assuming the underlying filtration of the problem satisfies 'the usual conditions'. But the proof of the theorem never explicitly mentions compactness or separability, and analyzing the steps missing from the argument, it doesn't seem they need these conditions either (my arguments seem to only require the set is closed, and the metric space arbitrary). Are these conditions needed in the proof, and if so, when are they utilized? Here is the proof, taken from the book:
Let $S_1(\omega) = A_K = \inf \{ t \geq 0: \text{either}\ X_t\ \text{or}\ X_{t^-}\ \text{is in}\ K \}$. Define $S_\eta$ for countable ordinals $\eta$ as follows: $$ S_{\eta + 1} = \inf \{ t \geq S_\eta: X_t\ \text{or}\ X_{t^-}\ \text{is in K} \} $$ $$ S_\eta = \lim_{\gamma \leq \eta} S_\gamma\ \text{if $\eta$ is a limit ordinal} $$ The argument used to prove lemma 74.4 shows that each $S_\eta$ is a $\{ \mathcal{F}_t \}$ stopping time. The process $X$ may approach $K$ at $S_\beta$, and `jump away at the last minute', in which case $S_{\beta + 1} = S_\beta$. But it can only make countably many such jumps away. (It is easy to prove that if the sum of an arbitrary sequence of non-negative terms is finite then only countable many terms can be nonzero.) We have $D_K(\omega) = S_{\delta(\omega)}(\omega)$ where $\delta(\omega)$ is the first - necessarily countable - ordinal such that $S_{\delta(\omega)}(\omega) = S_{\delta(\omega)+1}(\omega)$. The problem is that the number of countable ordinals, the number of possible values of $\delta(\omega)$, is uncountable (in much the same way that the number of finite ordinals is infinite). The probability measure $\mathbf{P}$ comes into play. Let $c_\eta = \mathbf{E} e^{-S_\eta}$, the infinum being over all countable ordinals. For $n \in \mathbf{N}$, we can find $\eta(n)$ such that $c_{\eta(n)} < c + 1/n$. Let $\eta(\infty)$ be the countable ordinal $\lim \eta(n)$. The $\eta(\infty)$ is independent of $\omega$, and, since $c_{\eta(\infty)} = c$, we have $S_{\eta(\infty)} = \sup S_\eta$ almost surely. It is now clear that $D_K$ is almost surely equal to the $\{ \mathcal{F}_t \}$ stopping time $S_{\eta(\infty)}$.