If $K$ is compact and $(f_k)\subseteq C(K)$ is dense, then $x_n\to x$ in $K$ iff $f_k(x_n)\to f_k(x)$ for each $k$

Assume that $(x_n)$ does not converg to $x$. Since $K$ is compact then $(x_n)$ has a subsequence say $(a_n)\subseteq(x_n)$ convergent to $a\neq x$. By continuity of any $f_k$ we have $f_k(a_n)\to f_k(a)$. So all we need now is to find $f_k$ such that $f_k(a)\neq f_k(x)$ to obtain a contradiction with uniqueness of the limit (among subsequences).

So assume that $f_k(a)=f_k(x)$ for all $k$. By the Tietze extension theorem there is continuous $g:K\to\mathbb{R}$ such that $g(a)=0$ and $g(x)=1$. The point is that $g(a)\neq g(x)$. And therefore it cannot be a limit of $f_k$ contradicting them being dense.

Hints: first show that $f(x_n) \to f(x)$ for all $x$. Then note that $f(y)=\frac {d(y,B(x,2\epsilon)^{c})} {d(y,B(x,2\epsilon)^{c})+ d(y,D(x,\epsilon))}$ defines an element of $C(K)$ where $D(x,r)$ denotes the closed ball with center $x$ and radius $r$. Note that $f$ is continuous, $0 \leq f \leq 1$, $f=1$ on $D(x,\epsilon)$ and $f=0$ on $B(x,2\epsilon)^{c}$. Now it should be easy to conclude (using the fact that $f(x_n) \to f(x)$) that $x_n \in B(x,2\epsilon)$ for $n$ sufficiently large. Hence $x_n \to x$.

To prove that $\rho$ is equivalent to $d$ use that fact that $\sum_{k=N}^{\infty} a_n (|f_k(x)-f_k(y)|\wedge 1) \leq \sum_{k=N}^{\infty} a_n$ can be made small independently of $x$ and $y$ by choosing $N$ large enough. From this you can see that a $\rho(x_n,x) \to 0$ iff $d(x_n,x) \to 0$.