This is from baby Rudin problem 7.9:
Let $f_n$ be a sequence of continuous functions which converges uniformly to a function $f$ on a set $E$. Prove that:
$$\lim_{n\rightarrow \infty} f_n(x_n)= f(x)$$
for every sequence of points $x_n \in E$ such that $x_n \rightarrow x$ for $x \in E$.
What I've tried
We want to show:
$$|f_n(x_n) - f(x)| \rightarrow 0$$
By the hypothesis on $x_n$:
$$|f_n(x_n) - f(x)| = \left|f_n\left(x + o(1)\right) - f(x)\right| = \left| f_n\left(x \right) + o(1) - f(x) \right|$$
Where the second equality comes from the continuity of $f_n$. This clearly goes to $0$ as $n\rightarrow \infty$. Nowhere have I used the uniform convergence of $f_n$ to $f$. What did I miss?
I have seen Why do I need uniform convergence here? but my argument is different.
I'm not a huge fan of the $o(1)$ notation being used here, and this may be hiding where the mistake lies. It is true that for any fixed $n$ you have $f_{n}(x + o(1)) = f_{n}(x) + o(1)$ (using your notation). However, this may not be true for all $n$ sufficiently large. In fact, saying that it is true for all $n$ sufficiently large is virtually the definition of uniform convergence! Without uniform convergence, what can happen is that as $n$ gets larger, the $o(1)$ quantity converges more and more slowly to zero, and in the limit may no longer converge to zero at all!