TITLE: Cardinality of the Euclidean topology and the axiom of choice
QUESTION [11 upvotes]: It is relatively straightforward to prove that the Euclidean topology has the same cardinality as the space itself.
I have sketched a proof below.
The proof seems to rely quite heavily on the axiom of choice.
I'm perfectly fine with (countable) dependent choice — I want to be able to define sequences recursively, for example — but one doesn't usually need more choice for topology in Euclidean spaces.
Is dependent choice enough to show that the topology of $\mathbb R^n$, $n\geq1$, has the same cardinality as $\mathbb R^n$?
If yes, how to prove it?
If not, how much do we know about the cardinality of the topology?
It seems that the cardinality of the topology lies between $\mathfrak{c}$ and $2^{\mathfrak{c}}$, but are there tighter bounds?
Proof of equal cardinalities with choice:
Let $T$ be the topology of $\mathbb R^n$ and $B$ a countable basis for it (rational balls, for example).
Since $|2^B|=|\mathbb R^n|$, it suffices to find an injection $f:\mathbb R^n\to T$ and a surjection $g:2^B\to T$.
These together will establish $|\mathbb R^n|\leq|T|\leq|2^B|=|\mathbb R^n|$ and the conclusion follows.
(Alternatively, the surjection $g$ has an injective right inverse, and we thus have injections both ways between $\mathbb R^n$ and $T$. The existence of a bijection follows from the Schröder–Bernstein theorem.)
We can simply take $f(x)=B(x,1)$ and $g(A)=\bigcup_{U\in A}U$.
Injecting $\mathbb R^n$ to $T$ requires no choice and always $T\subset 2^{\mathbb R^n}$, so at least $\mathfrak{c}\leq|T|\leq2^{\mathfrak{c}}$ without AC.
Without choice, I can't compare cardinals so easily or produce injective right inverses for surjections, so my proof falls apart.

REPLY [12 votes]: You don't need choice here at all - every open set $U$ has a canonical description in terms of basic open sets, $code(U)=\{B_\eta: B_\eta\subseteq U\}$ (where $\{B_\eta: \eta<\omega\}$ is the set of all balls with rational radius and center). We clearly have $code(U)=code(V)\iff U=V$, and the set of codes has size continuum.