If $\hat{O}$ is compact, then $\frac{\hat{O}}{\hat{p}}$ is finite

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Let $\hat{O}$ be complete discrete valued field $K$'s valuation ring and $\hat{p}$ be the maximal ideal of $\hat{O}$.

"If $\hat{O}$ is compact, then $\frac{\hat{O}}{\hat{p}}$ is finite.

Pf: As an image of $\hat{O}$ under continuous surjection $\hat{O}\to\frac{\hat{O}}{\hat{p}}$, $k$ is compact. As $\hat{p}$ is open in $\hat{O}$, $\frac{\hat{O}}{\hat{p}}$ must be discrete... "

$\textbf{Q:}$ Why is $\frac{\hat{O}}{\hat{p}}$ is discrete obvious here? The reference is algebraic number theory by Taylor Frohlich Pg 87 3.30. I found the argument in milne by pulling back the points in $\frac{\hat{O}}{\hat{p}}$ to open covering of $\hat{O}$ and using compactness of $\hat{O}$ to deduce finiteness of $\frac{\hat{O}}{\hat{p}}$. The book did not even mention open mapping or closed mapping here. I am interested in how the book deduce the discreteness here.