Let $(M,g)$ be a spacetime. An open set $A\subset M$ is causally convex if no causal curve intersects $A$ in a disconnected set. If $(M,g)$ is strongly causal, then for any point $p\in M$ and neighborhood $U$ of $p$, there is a causally convex set $A$ such that $p\in A\subset U$. It might be convenient to assume that $A$ can be taken to be a convex normal neighborhood (CNN), i.e. $A$ can be covered by normal coordinates centered at any $q\in A$ and that any $r,s\in A$ can be connected by a unique geodesic in $A$. Is this possible in general?
The existence of such sets is nonchalantly declared on page 195 of Hawking and Ellis, and on page 60 of Beem et al.
I attemped a proof by a standard argument: Take a CNN $U$ of $p$, and a causally convex set $A$ with $p\in A\subset U$. Then take another CNN $V$ of $p$ with $V\subset A$. Then $V$ should be causally convex or some small modification is. This seems to make sense, because causal convexity is supposed to protect from global "almost" causal violations. But in the case of $\Bbb R^2$ Minkowski space we can see that this program fails. A CNN is any set that is convex in the linear algebra sense, in particular a square with "the flat part down." But this is not causally convex because you can have a timelike curve that passes through the bottom corner (+time pointing up), then leaves through the side, and comes back in near a corner at the top. But if you rotate so it's a diamond, you get a causally convex CNN.
So maybe a causal diamond like $I^+(r)\cap I^-(s)$ with $r\ll p$ and $p\ll s$ could work, where $r$ and $s$ are "close enough" to $p$. I'm not sure how the details play out.