I believe the answer is yes. (Please check this answer carefully, as this is rather outside my field. There may well be a much easier solution.)
In the paper "Borel Orderings" by Harrington, Marker and Shelah (Trans. AMS 310 (1988), 293-302, MR 0965754), the authors consider Borel (partial) orderings. A Borel ordering is said to be thin if there does not exist a perfect set of pairwise incomparable elements. Of course, in a total order there are no incomparable elements, so every Borel total order is thin. Corollary 3.2 states that a thin Borel order does not contain an $\omega_1$-chain. (The original reference given for this result is a 1982 paper of Harrington and Shelah, "Counting equivalence classes for co-$\kappa$-Souslin relations", MR 673790, text at ScienceDirect, for which I lack the appropriate subscription.)
Given this, suppose $\preceq$ is a Borel total order on the Polish space $X$ and $A \subset X$ is convex. Using Zorn's lemma, choose a subset $C \subset A$ which is well-ordered by $\preceq$ and cofinal in $A$ (see Well-ordered cofinal subsets). If $C$ is uncountable then it contains a subset with the order type of $\omega_1$, which is forbidden by the Harrington–Shelah result. So $C$ is countable. Fix some $x_0 \in A$; then we have $A \cap [x_0,\infty) := \{x \in A: x \succeq x_0\} = \bigcup_{y \in C} [x_0,y]$. (The $\subseteq$ direction is because $C$ is cofinal, and the $\supseteq$ direction is because $A$ is convex.) This is a countable union, and each closed interval $[x_0,y]$ is Borel. So $A \cap [x_0, \infty)$ is Borel. By applying the same argument to the reverse ordering, we also get that $A \cap (-\infty, x_0]$ is Borel. The union of these two sets is $A$, so $A$ is Borel.