Let $(\mathbb{N},\tau)$ be a topological space, where $\tau=\{\emptyset, \mathbb{N}, \{0\},\{0,1\},\{0,1,2\},\dots\}$ I'm stuck here:

Let $(\mathbb{N},\tau)$ be a topological space, where
$$\tau=\{\emptyset, \mathbb{N}, \{0\},\{0,1\},\{0,1,2\},\dots\}$$
a) Prove that is not compact.
b) Prove that every continuous function $f: (\mathbb{N},\tau) \to \mathbb{R}$ is constant and hence bounded.

Part a) is easy: obviously $\tau$ without $\mathbb{N}$ and $\emptyset$ is a cover for $(\mathbb{N},\tau)$. Suppose that there's a finite subcovering for $\mathbb{N}$. Then we take the biggest set to see that the next element is not in this subcover and that it doesn't cover $\mathbb{N}$, so it's not compact.
Part b) is what I cannot see. How can I show that every continuous real valued set with domain that topological space is always constant?
Thanks for your time.

If $a=f(m)\ne f(n)=b$, and $f$ is continuous, then the sets
$$
f^{-1}\Big(-\infty, \frac{a+b}{2}\Big), \quad
f^{-1}\Big(\frac{a+b}{2},\infty\Big)
$$ 
and open, disjoint and non-empty. No such sets in $\tau$.
