$(f(x_n))_{n\in \mathbb{N}}$ converges for all continuous $f$. Does $(x_n)_{n \in\mathbb{N}}$ converge? Let $(S, d)$ be a complete metric space and $(x_n)_{n\in \mathbb{N}}$ a sequence in $S$. If $(f(x_n))_{n\in \mathbb{N}}$ converges for all continuous $f:S\to\mathbb{R},$ does it follow that $(x_n)_{n\in \mathbb{N}}$ converges with respect to $d$?
Intuitively, I think the implication is true, but I can't prove it (can't find a counterexample either).
Any ideas?
 A: This is true. In particular, consider any sequence $x_n$ such that all continuous functions $f:S\rightarrow \mathbb R$ converge. We will start by showing that if $x_n$ has a limit point $y$, then it converges to $x$. In particular, the condition of being a limit point may be expressed as
$$\liminf_{n\rightarrow\infty}d(x_n,y)=0$$
however, since $x\mapsto d(x,y)$ is a continuos map the limit of it applied to $x_n$ must exist and therefore coincide with the limit infimum, giving
$$\lim_{n\rightarrow\infty}d(x_n,y)=0$$
meaning $x_n$ converges to $y$. 
Next, we will show by contradiction that $x_n$ has a limit point. Suppose it didn't. Note two things: First, the image of the sequence $X=\{x_n\}$ is infinite, as no point can appear infinitely often. Second, any subset $S\subseteq X$ is closed, as any point on the boundary of $S$ is either in $S$ or is a limit point of $x_n$. Knowing this, let us take any subset $S$ of $X$ such that both $S$ and $X\setminus S$ are infinite.
At this moment, you could apply the Tietze extension theorem to a continuous function $f$ which is $1$ on $S$ and $0$ on $X\setminus S$ and note that $f(x_n)$ would not converge. If you want to work constructively, then enumerate $S$ as $\{s_1,s_2,\ldots\}$. Define $\varepsilon_n=\min(2^{-n},\inf\{d(s_n,y):y\in X\setminus \{s\}\})$. This is always positive since $X\setminus \{s\}$ is closed and doesn't contain $s$. Now, define the function
$$f(x)=\sum_{n=1}^{\infty}\max\left(1-\frac{1}{\varepsilon_n}d(x,y_n),0\right)$$
The reader may check that this is continuous - the trick to showing this is noting that, for any $x$, there is some ball around it in which only finitely many of the summands are non-zero. Moreover, the sum is zero on $X\setminus S$ and one on $S$, so $\lim_{n\rightarrow\infty}f(x_n)$ fails to exist. This is the contradiction we were seeking. Thus we conclude that $x_n$ must have a limit point and therefore must converge.
