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](http://www.ams.org/journals/tran/1988-310-01/S0002-9947-1988-0965754-3/) (Trans. AMS 310 (1988), 293-302, [MR 0965754](http://www.ams.org/mathscinet-getitem?mr=965754)), 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](http://www.ams.org/mathscinet-getitem?mr=673790), [text at ScienceDirect](http://www.sciencedirect.com/science/article/pii/S0049237X09705087), 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 https://mathoverflow.net/questions/82642/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.