Separating the étalé space

We briefly recall the construction of the étalé space of a sheaf. Given a sheaf $F$ over a space $X$, the associated étalé space is a topological space $E$ which, as a set, is the disjoint union of the stalks of $F$, with the topology induced by the basis $B(s, U) = \{s_x : x\in U\}$, where $U$ is an open set in $X$, $s\in F(U)$ and $s_x$ is the germ at $x$ of the element $s$.

Given a topological space $X$, consider the sheaf of real-valued continuous functions on $X$, which we denote $C$. It is more or less easy to see that the associated étalé space is not Hausdorff iff the space $X$ has the following property $(\star)$:

There exist continuous functions $f,g:U\to\mathbb{R}$, where $U$ is an open set, and a point $x\in U$ such that the germs of $f$ and $g$ at $x$ are different, but nevertheless for any open neighborhood $V\subseteq U$ of $x$ there exists an open set $W\subseteq V$ such that $f|_W = g|_W$.

For instance, $\mathbb{R}$ has property $(\star)$, and it’s easy to see that $(\star)$ is inherited by subspaces and products which contain any $(\star)$ factor. This, for instance, shows that subspaces of $\mathbb{R}^n$ are spaces satisfying this property.

Is there any characterization of all topological spaces for which the étalé space associated with the sheaf of real-valued continuous functions is not Hausdorff?

Free subgroups of the group of homeomorphisms

Let $n\geq 2$. Prove that the group of homeomorphisms of the $n$-dimensional ball that fix its boundary contains a copy of the free group on $2$ generators.

Spooky geometry III

Construct a path-connected metric space $X$ and a discontinuous function $f:X\to \Bbb{R}$ such that $f\circ \sigma$ is continuous for any continuous path $\sigma:[0,1]\to X$.

Spooky geometry II

Find a compact metric space that does not embed in $\Bbb{R}^n$ for any $n$.

Connect the dots

Construct a countable, connected topological space with at least two points, satisfying the highest separability axiom you can.

Notice that if two points can be separated by a continuous function, then a connected space with more than one point is uncountable, so your space cannot be $T_{3 \frac 12}$ (or higher).

Complements of closed sets

Construct  two homeomorphic closed subsets $A,B$ of some Euclidean space such that their complements are not homeomorphic, but show that if $A,B$ are closed homeomorphic subsets of $\Bbb R^{n+m}$ such that $A$ lies in $\Bbb R^n\times 0$ and $B$ lies in $0\times\Bbb R^m$, their complements in $\Bbb R^{n+m}$ are homeomorphic.

Self isometries of a compact metric space

Consider an isometry from a compact metric space to itself, this is simply a function that preserves distances. Notice that this implies that the function is continuous and moreover that it is injective. What we have to prove is that such a function is also surjective. Observe that this implies that the function is indeed a homeomorphism.

A consequence of this fact is that the category that has compact metric spaces as objects and isometries as arrows enjoys a Cantor-Schröder-Bernstein-like property.