$A$ be a subset of a Hilbert Space $H\ni \|x\|=1\forall x\in A$ and there is an $\epsilon\ni\|x-y\|\ge\epsilon\forall x,y\in A,x\neq y$ I need to know whether $A$ is finite or not.
Intuitively, It is clear that the set has not to be finite, the points satisfying these conditions are points on surface of a sphere.
If $H$ is finite dimensional then $A$ must be finite - else keep adding points to $A$ to maintain the $\|x-y\|$ property until $A$ is maximal. Then the set of open balls of radius $\epsilon$ around elements of $A$ is an open cover of the unit sphere.
If $H$ is infinite-dimensional we should be able to use the Riesz Lemma to construct a counter example for any non-compact $H$:
Let $H$ be an infinite dimensional Hilbert space, and pick some $x \in H$ with $\|x\|=1$. Let $A = \{x\}$. Let $B= $span$(A)$. Then $B$ is closed, so by the Riesz Lemma, we can chose $x_1 \in H$ with $\|x_1\|=1$ and $\|x-x_1\|>\epsilon$. Add $x_1$ to $A$ and repeat - since $H$ is infinite dimensional, the process does not terminate.
Note that all of this would work in a regular normed space.
In a Hilbert space, explicitly, you can take any orthonormal basis - since $$\begin{align}\|x-y\|^2&=\|x\|^2 + \|y\|^2 = 2\end{align}$$by Pythagoras.