$f$ defined on discrete metric space is a constant function? I saw the following statement:

Let $\mathbb{Z}^+$ be the set of positive integers and let $d$ be the discrete metric. Every real-valued function defined on $\mathbb{Z}^+$ is continuous.

I thought it wasn't too difficult. We can try $\delta = 0.7$. But then I came up with the following argument, which I suspect to be wrong? (especially the italic part, I seem to be making some logic error here and confusing myself)

To show that $f$ is continuous at $x_0$, for given $\epsilon>0$, we need to find $\delta >0$ such that
$$d(x,x_0)<\delta \Longrightarrow |f(x)-f(x_0)|<\epsilon \text{ for } x \text{ in the domain}$$
If we let $\delta = 0.7$, then in fact we would have $d(x,x_0)=0$. But this implies that $x=x_0$. As a result, in fact $f(x)=f(x_0)$.

The result of the argument looks really strange to me. First, it seems like we cannot actually choose $x$ to be different from $x_0$ because if we do, the distance between them will always be $1$, and thus forcing the function to be discontiuous? Second, does $f$ really have to be constant function?
I am pretty bad with topology but this argument just doesn't look right to me. And I would like to know where I got it wrong.
 A: The idea behind this proof is that continuous functions send convergent sequences to convergent sequences, limits being sent to limits. So, if we have a sequence $(p_n)$ in a metric space $M$ equipped with the discrete metric such that $p_n\to p\in M$, we have to consider the sequence $(fp_n)$ in $\Bbb R$ and assess its convergence. Since $p_n \to p$, we see that this sequence must eventually be constant. Hence, the sequence $(fp_n)$ must eventually be the constant sequence $fp,fp,\ldots$. Clearly for each $\varepsilon \gt 0$, it follows that there exists $N\in \Bbb N$ such that $n\in \Bbb N$ and $n\ge N$ imply that $\vert fp_n - fp\vert = 0 \lt \varepsilon$. Hence any real-valued function defined on a metric space equipped with the discrete metric must be continuous.
The classic $\varepsilon,\delta$ proof is pretty tricky to work in this case, and the equivalent definition of convergence defined by the mapping of convergent sequences to convergent sequences is cleaner and more intuitive to understand.
A: There is actually nothing wrong in your argument. Let's complete it and (hopefully) that helps you see it in a different way. 
For any $\epsilon >0$, you choose $\delta = 0.7$, and you know that $d(x,x_0)<0.7$ implies $x = x_0$ and trivially $f(x) = f(x_0)$. Thus you have proved: 

If $d(x, x_0) <\delta$ then $|f(x) - f(x_0)|<\epsilon$. 

And this implies that $f$ is a continuous function by definition. Since $f$ is arbitrary, we conclude that any function $f : X\to \mathbb R$ is continuous. 
I think may be the choice of $\delta = 0.7$ is bugging you. Note that if $f$ is a constant function, then $\delta$ can be any number since in the statement

If $d(x, x_0) <\delta$ then $|f(x) - f(x_0)|<\epsilon$,

the conclusion "$|f(x) - f(x_0)|<\epsilon$" is always satisfied. So if you really have a constant function, there is just no need to choose $\delta$ to be small. Put it another way, the choice of $\delta = 0.7 <1$ is to force the implication $d(x, x_0)<\delta \Rightarrow x=x_0$.  
