In other words, does the conclusion still follow?
Let $(f_n)$ be a sequence of functions defined on $A \subseteq \mathbb{R}$
I know that in the case of uniform convergence, it works because we can find an $N \in \mathbb{N}$ that works $\forall x \in A$
All I need to disprove it is a counterexample, like
$f_n(x) = \dfrac{nx}{1+x^2}$, with a functional limit of
$\dfrac{nx}{1+nx^2} \cdot \dfrac{1/n}{1/n}=\dfrac{x}{\dfrac{1}{n}+x^2}$, as $n \rightarrow 0$, it reduces to $\dfrac{1}{x}$ Obviously, on the interval $(0,1)$ it is not uniform convergent because
$|f_n(x)-f(x)|=|\dfrac{1}{x+nx^3}|<\epsilon$ means that $\dfrac{1-\epsilon x}{\epsilon x^3}\geq N$, where $N$ depends on $x$.
So, I'm on the right track here?
You just need to pick a non-continuous function that you can converge to pointwise with continuous functions and you'll have your counterexample.
May I suggest a step function that has a converging sequence of pointwise linear functions? You can not only arrange for pointwise convergence, but for convergence within a finite number of terms for any given point!