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Take the vertices of an $n$-simplex, and apply the permutation. If this turns the simplex inside out, then the permutation is odd; otherwise it is even. (This is basically equivalent to the comment by markvs, if one thinks of the determinant geometrically.)
if $1/9 < f(1/4) < 2/9$ then choose $I_2$ to be either $[0,1/4]$ or $[1/4,1/2]$. If $I_0 = [1/2]$ instead, choose $I_1$ by looking at $f(3/4)$, and whether it is $\leq 7/9$ or $\geq 8/9$. Continue on in this way, and in the end, take $x = \bigcap_{n \in \mathbb N}I_n$. Then $(x,f(x))$ is in your Cantor space.
Wait, here's a much simpler idea for a proof. Let $f$ be a continuous function $[0,1] \rightarrow [0,1]$. We're going to choose an infinite decreasing sequence of intervals in the domain. Let $I_0 = [0,1]$. To choose $I_1$, look at $f(1/2)$: if $f(1/2) \leq 1/3$, then choose $I_1 = [0,1/2]$, if $f(1/2) \geq 2/3$, then choose $I_1 = [1/2,1]$, and if $1/3 < f(1/2) < 2/3$, then choose $I_1$ to be either $[0,1/2]$ or $[1/2,1]$ (it doesn't matter which). Supposing $I_1 = [0,1/2]$, choose $I_2 = [0,1/4]$ if $f(1/4) \leq 1/9$, choose $I_2 = [1/4,1/2]$ if $f(1/4) \geq 2/9$, and . . .
cannot be functional. Similarly, you can argue that the graph of every continuous function meets $K_2$, and $K_3$, etc. So now fix some particular $f$. The graph of $f$ is compact, and so the sets $\mathrm{Graph}(f) \cap K_n$ form a decreasing sequence of compact sets. So their intersection is nonempty.
Nik, here is an idea for proving your example works. Write your Cantor space as the intersection of an infinite decreasing sequence $K_0 \supseteq K_1 \supseteq K_2 \dots$ of closed sets. $K_0$ can be the whole square. $K_1$ should look like two thick strips of positive slope, with a gap between them. $K_2$ should be 4 thick strips, even slopier, with one pair contained in each of the strips from $K_1$, etc. The graph of every continuous function meets $K_1$, because any path from the left-hand side to the right-hand side of the square has to snake between the two strips, and therefore . . .
@YCor: There are two well-known proofs without CH, neither of which assumes the negation of CH. The first (Frolik's, with some help from Kunen) examines a partial order on ultrafilters, now known as the Rudin-Frolik order, to show $\mathbb N^*$ is not homogeneous. The second (due to just Kunen) constructs something called a "weak $P$-point" using just ZFC. This feels much more like a ZFC extension of Rudin's proof. But I should mention that, by a theorem of Shelah, ZFC cannot prove that $\mathbb N^*$ contains any $P$-points. So strictly speaking, Rudin's proof cannot work with just ZFC.
This is a very interesting construction. For what it's worth, I think you can modify things slightly to make it work in the Cohen model. (I don't know how interesting this would be to you, and whether it's worth the trouble of writing down the details.) This shows that the existence of such a space is consistent with $\mathfrak{d} > \aleph_1$, but it does not answer your question about what happens when $\mathfrak{b} > \aleph_1$.
@MathieuBaillif: I agree with your first sentence, but not your second. I think what you meant is that a scattered Hausdorff space has Lindelöf number at least as big as its cofinality. For example, $\omega_1+1$ is compact.
