Continuous bounded functions determine the topology

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I wonder wether this is true and how to prove it:

Let $X$ be locally compact topological Hausdorff space and let $\text{C}_b(X)$ denote the bounded continuous functions thereon. Let furthermore $(x_\lambda)_\lambda$ be a net in $X$, $x\in X$. The following are equivalent:

  1. $x_\lambda\rightarrow x\in X$
  2. $\forall f\in \text{C}_b(X) \colon f(x_\lambda)\rightarrow f(x)$

It feels like this should be a Corollary to Urysohn's lemma and could also be true already for smaller function algebras, but I cannot see through it.

Thanks in advance!

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This is indeed true and follows from the fact that $X$ is a Tychonoff space: for all $x \in X$ and $C \subseteq X$ closed such that $x \notin C$ there exists a continuous $f: X \to [0,1]$ such that $f(x) = 1$ and $f[C] = \{0\}$.

Now 1 always implies 2 by continuity of all $f \in C_b(X)$. If we assume 2. and to show 1., let $O$ be any open neighbourhood of $x$. Find $f: X \to [0,1]$ vanishing on $X\setminus O$ and with $f(x) =1$. Then 2. says that $f(x_\lambda) \to f(x)$, so that $f(x) \in (\frac{1}{2},1]$ which means that $f(x_\lambda)$ should eventually be in $(\frac{1}{2},1]$ as well, which implies that $x_\lambda \in O$ eventually. As $O$ was arbitrary the net converges to $x$.