When is the squeeze theorem true? Here is the Squeeze Theorem in $\mathbb{R}$:

Let $(a_n)$, $(b_n)$ and $(c_n)$ be sequences taking their values in $\mathbb{R}$. Let $x \in \mathbb{R}$. Assume that:
  
  
*
  
*$\forall n \in \mathbb{N}, \ \ a_n \leq b_n \leq c_n$;
  
*$\lim \limits_{n \to + \infty} a_n = \lim\limits_{n \to + \infty} c_n = x$.
  
  
  Then $\lim \limits_{n \to + \infty} b_n = x$.

This theorem is true is one replaces the occurences of $\mathbb{R}$ above by $\overline{\mathbb{R}}$, $\mathbb{R}^n$, $\mathcal{C}_b (\Omega)$ (where $\Omega$ is an open set), $\mathbb{L}^p (\Omega, \mu)$ (where $(\Omega, \mu)$ is a measured space and $p \in [0, + \infty]$), and even in $\mathcal{P} (\mathbb{R})$ (see this related question). However, the proofs I know of these facts have some points in common, but also some individual ingredients.
Is there a general sufficient condition which would ensure that a topological space with a partial order satisfies the Squeeze Theorem, and apply to all examples above? Are there some not too contrived examples of spaces for which the Squeeze Theorem fail?
 A: This may not quite answer your question, but it was too long for a comment, and worth pointing out, IMHO :
Many of the function/sequence spaces where this proof works are all Banach lattices. ie. They are ordered normed linear spaces, and the norm respects the order structure.
For such spaces, a functional $f:E \to \mathbb{C}$ is called positive, if $x \geq 0$ implies that $f(x) \geq 0$. In particular, if $a_n \leq b_n \leq c_n$ for all $n$, then
$$
f(a_n) \leq f(b_n) \leq f(c_n) \quad\forall f \text{ positive linear functional }
$$
Hence, by the squeezing principle in $\mathbb{R}$, it follows that
$$
f(b_n) \to f(x) \quad\forall f \text{ positive linear functional }
$$
Now, there is a theorem that say that the positive linear functionals generated the (continuous) dual space. In other words,
$$
E^{\ast} = \{f-g : f,g \text{ positive linear functionals }\}
$$
Hence (by the Banach lattice version of the Hahn-Banach theorem), it follows that $b_n \to x$ in $E$
A: At least, for metric spaces with a partial order I found some generalizations of the squeeze theorem. They are useful to deduce the cases $\mathbb{R}$, $\mathbb{R}^n$, $\mathcal C_b(\Omega)$ and $\mathcal L_p(\Omega)$.
First, I came up with the following proposition:

Proposition. Let be $(X,d)$ a metric space, $a\in X$, $(x_n)$, $(y_n)$ sequences in $X$ and $A_n\subseteq X$ for $n\in\mathbb{N}$ with the following properties:
  
  
*
  
*$x_n,y_n\in A_n\quad\forall n\in\mathbb{N}$
  
*$\operatorname{diam} A_n=\sup_{x,y\in A_n} d(x,y)\rightarrow 0$
  
*$x_n\rightarrow a$
  
  
  Then we also have $y_n\rightarrow a$.

One can easily check this by elementary estimations. Now we conclude the general squeeze theorem for metric spaces $(X,d)$ endowed with a partial order $(\preceq, X)$ with respect to the metric $d$, that means we have $d(x,y)\leq d(x,z)$ for all $x\preceq y\preceq z$.

General squeeze theorem. Le be $(X,d)$ a metric space, $(\preceq,X)$ a partial order , $a\in X$ and $(x_n)$, $(y_n)$, $(z_n)$ sequences in $X$ with
  $$x_n\preceq y_n\preceq z_n$$
  for each $n\in\mathbb{N}$ and $x_n,z_n\rightarrow a$. Then it follows that $y_n\rightarrow a$.

Proof. Define $A_n=\{y\in X\ \vert\ x_n\preceq y\preceq z_n\}$. Then we have $x_n,y_n\in A_n$ for each $n\in\mathbb{N}$ and
$$\operatorname{diam}A_n=d(x_n,z_n)\rightarrow 0.$$
Thus the theorem follows from the proposition. $\square$
