Let $X$ and $Y$ be metric spaces. Assume that $Y$ is a discrete metric space and that $f : X \longrightarrow Y$ is a contraction. What can be concluded about $f\ $?
What I observe is that if there exist $x, y \in X$ with $x \neq y$ such that $f(x) \neq f(y)$ then we have $$1 = d(f(x),f(y)) \leq c\ d (x,y)$$ for some $0 \lt c \lt 1.$
Now let $x_0 \in X.$ Let there exist an $y \in X$ such that $f(y) \neq f(x_0).$ Then since $Y$ is discrete it follows that $d(f(x_0),f(y)) = 1.$ Let us choose $0 \lt \varepsilon \lt 1.$ Then by continuity of $f$ (any contraction map is Lipschitz and hence uniformly continuous) there exists $\delta \gt 0$ such that for all $x \in B(x_0,\delta)$ $$d(f(x),f(x_0)) \lt \varepsilon \lt 1.$$ But then by the discreteness of $Y$ it follows that $f(x) = f(x_0),$ for all $x \in B(x_0,\delta).$ So $f(x) \neq f(y),$ for all $x \in B(x_0,\delta).$ But this shows that $d(f(x),f(y)) = 1,$ for all $x \in B(x_0,\delta).$ Since $f$ is a contraction map it follows that for all $x \in B(x_0,\delta)$ $$1 = d(f(x),f(y)) \leq c\ d(x,y)$$ for some $0 \lt c \lt 1.$
Does it lead to any contradiction?