The answer is no. Assume there is such a well-order; then some downward-closed subset $D$ (which might be an initial segment, or might be all of $\mathbb{R}$) will be of order type $2^{\aleph_0}$, i.e., the least ordinal of continuum size. Being a subset of $\mathbb{R}$, $D$ contains a countable subset that is dense in $D$ (see for example here if you are in need of proof), say $x_1 \lt x_2 \lt \ldots$ according to how the elements appear in the well-order. Since there is no countable cofinal sequence in the order type $2^{\aleph_0}$, it must be that some proper initial segment contains all the $x_i$. This initial segment is both closed and dense, hence is all of $D$. Contradiction.

(Edited after Ramiro pointed out a glitch. Hopefully okay now.)