Example of Hausdorff space $X$ s.t. $C_b(X)$ does not separate points? We know the Stone-Weierstrass theorem for locally compact Hausdorff spaces (LCH) which states the following:

Theorem: Suppose $X$ is LCH. A subalgebra $\mathcal{A}$ of $C_0(X)$ is dense if and only if it separates points ($\forall x,y \in X : x\neq y \implies \exists f \in \mathcal{A}: f(x) \neq f(y)$) and vanishes nowhere $(\forall x \in X \exists f \in \mathcal{A} : f(x) \neq 0$) and is closed under complex conjugation1. 

It's also easy to show that $C_b(X)$, the continuous and bounded functions on a topological space $X$, is a Banach space2. It would be obvious to try and show a variant of Stone-Weierstrass for $C_b(X)$ where $X$ is merely Hausdorff, and since it's obviously dense in itself, for such a theorem to exist it would be required that $C_b(X)$ separates points (that it vanishes nowhere is clear: it contains the constant functions). 
I've tried to come up with an example of a Hausdorff space where $C_b(X)$ fails to separate points (this could for example be a space where every non-constant continuous function is unbounded), but to no avail. So does anyone happen to have an instructive example lying around?
PS: It's an interesting exercise to prove that given a LCH space $X$ the continuous functions with compact support, $C_{00}(X)$, separates points and vanishes nowhere. You get to use many theorems from topology in the process of constructing a continuous function such that $f(x)=1$ and $f(y)=0$ given distinct points $x,y \in X$ [Hint: Find a good compact subset and use Urysohn's lemma].
[1]: In the case of real-valued functions this is essentially a no-op, so including it in the theorem statement doesn't hurt.
[2]: [Car00] shows in Lemma 10.8  that $B(X)$, the set of bounded functions on a set $X$ is a Banach space, and it's an immediate corollary from Thm 10.4 that $C_b(X)$ is closed in $B(X)$ for $X$ a metric space. Of course this  generalises readily to the case of $X$ a topological space.
 A: I'll just point out that boundedness is not the obstruction: any space that admits a nonconstant real-valued continuous function admits one which is bounded, and if $C(X,\mathbb{R})$ separates points then so does $C_b(X,\mathbb{R})$.  This is easy to see: if $f$ is continuous and $f(x_1)=a < b = f(x_2)$, then $g(x) = (f(x) \vee a) \wedge b$ is continuous and bounded and also has $g(x_1)=a<b=g(x_2)$.
However, continuous functions with compact support may not be enough if $X$ is not locally compact, even if it is metrizable.  For instance, it's a nice exercise to check that if $X$ is an infinite-dimensional Banach space, the only real-valued continuous function on $X$ with compact support is the zero function, i.e. in your notation $C_{00}(X) = \{0\}$.  (I think most authors call this space either $C_0(X)$ or $C_c(X)$.)
A: I recommend this short 1971 note of T.E. Gantner.  He first proves:
Theorem: Each regular space $Z$ can be embedded as a subspace of a regular space $Q(Z)$ such that every continuous, real-valued function on $Q(Z)$ is constant on $Z$.
He applies the theorem as follows: take $X_0$ to be a one-point space, and inductively, for all $n \in \mathbb{N}$, define $X_{n+1} = Q(X_n)$.  Let $X = \lim_{n \rightarrow \infty} X_n$, endowed with the direct limit (or final) topology.  Then $X$ is an infinite regular space on which every continuous real-valued function is constant.
A: Looking at the table at the back of Counterexamples in Topology, the first space which is Hausdorff but not Urysohn (which in that book means $C(X)$ doesn't separate points) is the relatively prime integer topology. That is, the topology on $\mathbb{N}\setminus \{0\}$ with basis all sets of the form $\{b+na|n \in \mathbb{N}\}$ for $a$, $b$ coprime.
Hausdorff is easy: if $k, l \in \mathbb{N}\setminus \{0\}$, take a prime $p$ larger than both and consider the open neighbourhoods $\{k+np \}, \{l+np \}$.
$C(X)$ fails to separate points because any two basic open sets have nondisjoint closures, since the closure of  $\{b+na \}$ contains all multiplies of $a$.
