This question is from section Product Topology by Wayne Patty.
Example 11: For each i $\in \mathbb{N}$, let $Y_i =\mathbb{R}$. Recall that $\mathbb{R}^{\omega}$ denotes the cartesian product $\prod_{i\in \mathbb{N}}Y_i$. Let T denote the usual topology on $\mathbb{R}$, and let U denote the box topology on $\mathbb{R}^{\omega}$. Define a function f: $\mathbb{R} \to \mathbb{R}^{\omega}$ as follows : For each x$\in \mathbb{R}$ and each n$\in \mathbb{N}$, let f(x) be the sequence whose value at n in x. Then for each i $\in \mathbb{N}$, $\pi_{i} \circ f$ is continuous but f is not continuous.
Solution: Let i$\in \mathbb{N}$. Then $\pi_i \circ f$ is the identity and it is continuous.
In order to show that f is not continuous, we exibit a member B of U such that $f^{-1}(B) $ doesn't belong to T. For each i $\in \mathbb{N}$, let $B_{i}=(-1/i , 1/i) $ and let B=$\prod_{i\in \mathbb{N}} B_i$. Let $B\in U$. The proof that $f^{-1}(B) $ doesn't belongs to T is by contradiction. Suppose $f^{-1}(B) \in T$. Then since $0\in f^{-1} (B)$, there is an open interval (a,b) such that 0$\in (a,b)$ and (a,b)$\subseteq f^{-1} (B)$. So, f((a,b))$\subseteq B$ and thus for each natural number n , $(\pi_n \circ f) ((a,b))=(a,b) \subseteq (-1/n ,1/n)$ . This is a contradiction.
How does in the line 4 of solution author deduced that $0\in f^{-1} (B)$
and how does $(\pi_n \circ f)((a,b)) \subseteq (-1/n, 1/n)$ is a contradiction.
Ad #1:
$$0\in f^{-1}(B)\Leftrightarrow f(0)\in B$$ What are $f(0)$? What is $B$?
Ad #2:
This is miswritten. It's not that $(\pi_n\circ f)((a,b))\subseteq\left(-\frac{1}{n},\frac{1}{n}\right)$ is a contradiction — it's that $$(a,b)\subseteq\left(-\frac{1}{n},\frac{1}{n}\right)$$ for all $n$ is a contradiction. Take $n$ large: do you see why?