Given a sequence of functions $f_n$ all continuous. If asked if there exists a subsequence $(f_{n_k})_{k \in \mathbb{N}}$ and continuous function $f$ in which $f_{n_k} \to f$ as $k \to \infty$, can I not just say $f:=f_1$ and $n_k=1$ for all $k$? Since then, $K$ bounded subset of $\mathbb{R}^n$:
$$\sup_{x \in K} |f_{n_k}(x)-f(x)|\to 0$$ as $k\to \infty$.
[Comment converted to answer:] That's not a subsequence. You can't repeat an index in forming a subsequence, you can only possibly skip it: $n_1<n_2<n_3<\cdots$