The question is: Let $f_n$ be a sequence of continuous functions defined on a compact set $K\subset \mathbb R$. Then $f_n$ converges to $f$ uniformly if and only if for any sequence $x_n\in K$ that converges to some $x\in K$, we have $f_n(x_n)\to f(x)$.
I could show the only if part, and I don't know how to show the if part. As far as I know:
If $f$ is continuous but not necessarily $f_n$, I could show the result is true.
If $K$ is not compact, then the result is not true.
Therefore my idea is to show that $f$ is continuous using the continuity of $f_n$. But I do not know how to proceed.
Note if $f_n$ is an equicontinuous sequence of functions on compact set $K$, and $f_n$ pointwise converges to $f$, then $f_n$ uniformly converges. See this for a proof.
Then it suffices to show $f_n$ is a equicontinuous sequence of functions. That is $\forall \epsilon>0$, there exist $\delta$, such that $\forall x,y\in K$, with $|x-y|<\delta$, we have $|f_n(x)-f_n(y)|<\epsilon,\forall n$.
Note $f_n$ is continuous on compact support hence uniformly continuous. Thus $f_n$ is a equicontinuous sequence of functions if and only if all the tail of $f_n$ is a sequence of equicontinuous functions.
Suppose $f_i, i\ge 1$ is not a equicontinuous sequence of functions, then there exists $\epsilon>0$ such that for all $\delta>0$, there exists $n\in \mathbb{N}$, $x=x_n,y=y_n\in K$, with $|x_n-y_n|<\delta$, but $|f_n(x)-f_n(y)|\ge\epsilon$. Note $f_i,i\ge n$ must not be a sequence of equicontinuous sequence of functions. In this way, we can extract two sequences of $x_n, y_n\in K$ such that $x_n-y_n\to 0$ but $|f_n(x_n)-f_n(y_n)|\ge\epsilon$ (Here we abuse the notation use $f_n$ to denote the extracted subsequence). But this will contradict to the assumption by passing to convergent subsequences.