Bounded sequence of functions implies convergent subsequence

Question

Here you can see my attempt at the proof. I am sure I did something wrong because my prof asked me to show it for rationals and I "somehow" showed it for all reals. I would appreciate it if someone can spot my mistake! Thanks

The reference to the theorem 3.6 is for Rudin "Principles of Mathematical Analysis":

Theorem 3.6: If $\{p_n\}$ is a sequence in a compact metric space X, then some subsequence of $\{p_n\}$ converges to a point of X

Answer 1

You chose a subsequence working for only one $x$. There is zero guarantee that this same subsequence will work for another.

More precisely, your compactness argument is as follows:

Let $x$ be a real number. The sequence $(f_n(x))_n$ is bounded, thus there exists an increasing $\varphi_x : \mathbb{N} \rightarrow \mathbb{N}$ such that $(f_{\varphi_x(n)}(x))_n$ is convergent.

But if $y$ is another real number, there is no reason why $(f_{\varphi_x(n)}(y))_n$ is convergent.