Proving a function on the discrete metric is continuous

Let $X$ be a set with the discrete metric and $ f:X \to Y$ a map of sets underlying metric spaces. Then $ f $ is automatically continuous.

Now, let's play with sequences. As long as you are in Hausdorff space, sequence is an important tool to play with limits.

In a discrete space $(X, d_{discrete}) $ , we can classify all convergent sequences. A sequence $(x_n) $ is convergent in $(X, d_{discrete}) $ iff it is eventually constant i.e $$\exists N\in \Bbb{N} , x_n=x_0 , \forall n>N$$

$ f:X \to Y$ is continuous at $x_0 \in X$ if for every sequence $(x_n) \subseteq X$ with $x_n \to x_0 $ implies $f(x_n)\to f(x_0) $ in $(Y, d')$.

As, $x_n\to x_0, \exists N\in \Bbb{N} , x_n=x_0 , \forall n>N$

Then, $f(x_n) =f(x_0), \forall n>N$

Implies, $f(x_n) \to f(x_0) $ in $(Y, d') $

Hence, $ f:X \to Y$ is continuous at $x_0$.

Since, $x_0 \in X$ is arbitrary, $f$ is globally continuous on $(X, d_{discrete}) $