"Let $X$ be contained in $\mathbb{R}^2$ such that the euclidian metric restricted to X equals the discrete metric. Prove that $X$ has at most 3 elements. What if $X$ is contained in $\mathbb{R}^3$? Generalize for $\mathbb{R}^n$. Could you imagine a normed vector space $E$ and an infinite subset $X$ contained in $E$ such that $x≠y$ in $X$ implies $d(x,y)=1$?"
It's the exercise 8 of the first chapter of the book "Espaços Métricos" (Metric Spaces) by Elon Lages Lima, in the section about the definition of metric space. Although it's very intuitive, I can't formalize a proof to it beyond proving the $\mathbb{R}^2$ and $\mathbb{R}^3$ cases with a lot of algebraic effort. The generalized case seems to be a proof by induction, but I'm not able to prove the $\mathbb{R}^{n+1}$ case assuming the $\mathbb{R}^n$ case, even trying some different ideias for treating the problem.
And I imagine that the last example the problem requires should be the $\mathbb{R}^\infty$, but I remember there is a difference between $\mathbb{R}^\infty$ and $\mathbb{R}^{(\infty)}$ but I don't know what is the difference neither which of these spaces I should use.
Here is the induction step:
Assume that the maximal solution for $X\subseteq\mathbb{R}^n$ is a set $|X|=n+1$ with a property that the center of the points in $X$, $c=\frac{1}{n+1}\sum_{x\in X}x$ is the only point equidistant from all points in $X$, and that distance is less than $\frac{\sqrt{2}}{2}$. This trivially holds for $n=1$. Let's say we have a set $X\subseteq\mathbb{R}^{n+1}$ satisfying your criteria. Since we can embed $\mathbb{R}^n$ into $\mathbb{R}^{n+1}$ we know that $|X|\geq n+1$. Take some $Y\subseteq X$, $|Y|=n+1$. Affine closure of $Y$, call it $A$, has dimension at most $n$, so by the inductive hypothesis it has dimenstion exactly $n$ (it can't be less) and there is a point $c'$ equidistant from all points from $Y$, with that distance being $d<\frac{\sqrt{3}}{2}$. Assume that WLOG $c'=0$ so that our $A$ is a vector subspace and so that we can talk about vectors. Let $B$ be a line containing $c'=0$ perpendicular to $A$ ($B$ is a one-dimensional vector sub-space). Take some $x\in X\setminus Y$ if it exists. Express $x$ as $x=a+b$ where $a\in A$ and $b\in B$. If $a\neq 0$ then there are some two points $x_1$ and $x_2$ from $X$ which are not equidistant from $a$, say $\text{dist}(x_1,a)=d_1\neq d_2=\text{dist}(x_2,a)$. We have that $\text{dist}(x_1,x)^2=d_1^2+|b|^2$ and similarly for $x_2$ which shows that $x$ isn't equidistant from $x_1$ and $x_2$ either. This means $a=0$, i.e. $x\in B$ $(\ast)$. From $\text{dist}(X,x)=1=\sqrt{d^2+|x|^2}$ we see that $|x|=\sqrt{1-d^2}>\frac{\sqrt{2}}{2}$. This last inequality shows that there can be only 1 such $x$ on the line $B$ because otherwise their distance would be greater than $2\cdot\frac{\sqrt{2}}{2}=\sqrt{2}>1$. This shows $|X|=n+2$ as wanted. It remains to show the property about the center. From $(\ast)$ we have that the center $c$ must be on the line $B$. Express $c$ as $c=t\cdot x$ where $t\in\mathbb{R}$. We have that $\sqrt{d^2+t^2(1-d^2)}=\text{dist}(X,c)=\text{dist}(x,c)=(1-t)|x|=(1-t)\sqrt{1-d^2}$. We derive that the new distance is $\frac{3\sqrt{1-d^2}-\sqrt{3-5d^2}}{2}$. This function is increasing for $d\geq\frac{1}{2}$ and for $d=\frac{\sqrt{2}}{2}$ it is equal $\frac{\sqrt{2}}{2}$, so indeed, the expression for $d<\frac{\sqrt{2}}{2}$ is less than $\frac{\sqrt{2}}{2}$ again.
The infinite case:
Embed $\mathbb{R}$ into $\mathbb{R}^2$, then embed that into $\mathbb{R}^3$, then embed that into $\mathbb{R}^4$, etc... The construct from above will give you an array of points where each point from the sequence has "only finitely many non-zero coordinates". So you can just take $E$ to be the set of all real sequences with only finitely many non-zero elements. Metric is the classical Euclidean.