Let $f : X \to \mathbb R$ be continuous and $A$ a dense subset of $X$. Let $$B_n = \{x \in X| − n < f (x) < n\}, n \in \mathbb N.$$ If for some $n_0$ the intersection $A \cap B_{n_0}$ is empty, prove that each $B_n$ is empty for $1 \leq n \leq n_0$
Definition: A mapping $f:X \mapsto Y$ is said to be continuous at the point $x$ of $X$ provided that for every $\epsilon > 0$ there is a $\delta >0$ such that $$y \in X, d_X(x,y) < \delta \implies d_Y(f(x),f(y)) < \epsilon$$
The mapping f is said to be continuous if it is continuous at every $x$ of $X$ .
Dense subset means $\overline{A} = X $
Clearly $B_n \subseteq B_{n_0}$ for such $n$, so we need only show that $B_{n_0}$ is empty.
Suppose not. Then there is some $x \in B_{n_0}$. Necessarily, $x \not\in A$.
Since $f$ is continuous, if we in particular take $\varepsilon := n_0 - |f(x)| > 0$ (this is true since $x \in B_{n_0}$), there is some $\delta > 0$ such that for all $y$ in$X$ such that $d_X(x,y) < \delta$, we have $d_\mathbb{R}(f(x),f(y)) < \varepsilon$.
But since $A$ is dense in $X$, there is, in particular, some $a \in A$ such that $d_X(x,a) < \delta$, hence $d_{\mathbb{R}}(f(x),f(a)) < \varepsilon$. But then \begin{align*}|f(a)| &\leq |f(x)| + |f(x)-f(a)| \\&= |f(x)| + d_{\mathbb{R}}(f(x),f(a)) \\&< |f(x)| + \varepsilon \\&= |f(x)| + n_0 - |f(x)| \\&= n_0,\end{align*} so $a \in B_{n_0}$, hence $a \in A\cap B_{n_0}$, but $A \cap B_{n_0} = \emptyset$, a contradiction.