Question

Assuming the axiom of choice, is there a well-ordering of the reals such that every initial segment is closed (for the usual topology)? If the continuum hypothesis helps we can also assume it.

An initial segment is a set of the form $\{x : x<y\}$ for some $y$, according to the well-ordering $<$.

Answer 1

No. There cannot be a strictly increasing $\omega_1$-sequence of closed sets in a topological space with a countable base. Their complements would be unions of open sets from the basis, and it is impossible to drop elements of a countable set uncountably many times.

Answer 2

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.)

Answer 3

No.

If there was such a well order, let $r$ be the first real for which the initial segment $I$ bounded by $r$ is uncountable. Since $I$ is separable (being a subspace of the reals) we can fix a countable $D \subseteq I$ dense in $I$. But $I$ has order type $\omega_1$ which is a regular cardinal (usinge some choice), so there would be an $s \in I$ that bounds $D$. This is already a contradiction because then the clousure of $D$ would be contained in the initial segment determined by $s$, which is a proper subset of $I$.