Every perfect set has cardinality $2^{\aleph_0}$?

It is well known that perfect sets in $$\mathbb{R}^n$$ are uncountable (e.g., baby Rudin states this). Recently I heard of this stronger result:

Every perfect set in $$\mathbb{R}^n$$ has cardinality $$2^{\aleph_0}$$.

This is easily proved if we assume the continuum hypothesis. However, this result does not rely on that. Is there a proof of this fact? And does this result hold for more general spaces (e.g., complete metric spaces)?

I understand that it's customary to show my effort here at math.SE, but honestly I have no idea how I should attempt at it...

Yes, it does not require the continuum hypothesis to prove.

Suppose $$P$$ is a perfect set. WLOG (if not, restrict to an appropriate closed interval) $$P$$ is bounded. Then we can find two disjoint closed subsets, $$l(P)$$ and $$r(P)$$, which are each also perfect. And we can iterate this process, building such things as $$r(l(r(P)))$$ and so on.

Now remember that the intersection of a descending sequence of closed and bounded sets is nonempty. This means that for every infinite binary sequence, the corresponding descending sequence of perfect sets has a point in the intersection. For example - identifying "$$r$$" with $$1$$ and "$$l$$" with $$0$$ - the sequence $$f=0,1,0,1,,...$$ yields the sequence $$l(P)\supseteq r(l(P))\supseteq l(r(l(P)))\supseteq r(l(r(l(P))))\supseteq...,$$ and we pick a point $$p_f$$ in the intersection of this chain of perfect sets.

Now just check that if $$f\not=g$$ we have $$p_f\not=p_g$$ (HINT: $$r(X)\cap l(X)=\emptyset$$ ...).

It might seem like we used the axiom of choice here, in two places:

• Choosing $$l(X)$$ and $$r(X)$$, for a perfect set $$X$$.

• Choosing a point in the intersection of a decreasing sequence of closed and bounded sets.

However, we don't actually need AC here:

• If $$X$$ is perfect, we can in fact show that there is a pair of rationals $$p such that $$X\cap (-\infty, p]$$ and $$X\cap [q,\infty)$$ are each perfect. But $$\mathbb{Q}^2$$ is well-orderable ...

• There is in fact an easily-definable choice function for nonempty closed sets. (HINT: if the set is bounded from below, just pick the smallest element; now do you see how to deal with the not-bounded-from-below case?) We could also require that $$l(X)$$ and $$r(X)$$ have diameter at most one half of that of $$X$$ (where the diameter of a set is the supremum of the distances between any two points in the set).