Since there is a homeomorphism of $\Bbb R $ taking $\Bbb Q $ to the dyadic rationals, that is, rational numbers where the denominator is a power of 2, (see here for an example of such a function), we can restrict out attention to finding a sequence of functions converging to infinity exactly at the dyadics.
Now for $ p, d \in \Bbb R $ we let $ h_{p, d} $ denote a hat function, i. e. a continuous function with $ h_{p, d}(p) =1$ and $ h_{p, d}(x) =0$ for $ x \notin [p-d, p+d]$.
Now define
$ f_n = \sum_{k \in \Bbb Z} n h_{k/2^n,2^{-n-2} }$
It is obvious that for dyadic x, $ f_n (x) \to \infty $ as $ n \to \infty $.
If $x$ is not dyadic, there are infinitely many $ n $ with $\{2^n x\} \in [1/4,1/2]$ (here the brackets denote the fractional part of the real number). To see this, consider the binary expansion of $ x $ and note that, since this expansion is not allowed to end in an infinite chain of zeros or in an infinite chain of ones, there are infinitely many integers $ n $ such that the $ n+1$-th digit after the dot is a $0$ and the $ n+2$-th is a $1$.
But with such a choice of $ n $, we get $ f_n (x)=0$ which shows that $ f_n (x )\not \to \infty $ as $ n \to \infty$.
Hint: $f(x) = |x|$ is continuous.
(Of course, the limits must exist for this to apply. For a counterexample, see the comment above made by coffeemath).
Best Answer
We note that if $x_n \to x$ and $g$ is continuous, then: $$ \lim_{n \to \infty} g(x_n) = g\left(\lim_{x \to \infty} x_n\right) $$ i.e. the limit and the function can be interchanged. The absolute value function $x \mapsto |x|$ is continuous. Therefore, in your case we have: \begin{align*} |f(x) - f(y)| &= \left|\left(\lim_{n \to \infty} f_n(x)\right) - \left(\lim_{n \to \infty} f_n(y)\right)\right| \\ &= \left|\lim_{n \to \infty} (f_n(x) - f_n(y))\right| \\ &= \lim_{n \to \infty} |f_n(x) - f_n(y)| \end{align*}