Why does a divergent correlation function (not length) imply lack of the order in the system? Consider a two-point correlation function defined as $$G_{ij}({\bf x},{\bf x}^\prime)\equiv \Big\langle\Big(\mathscr{O}_i({\bf x})-\big\langle\mathscr{O}_i({\bf x})\big\rangle\Big) \Big(\mathscr{O}_j({\bf x^\prime})-\big\langle\mathscr{O}_j({\bf x^\prime})\big\rangle\Big)\Big\rangle\tag{1}$$ $$~~~~~~~=\big\langle \mathscr{O}_i({\bf x})\mathscr{O}_j({\bf x}^\prime)\big\rangle-\big\langle \mathscr{O}_i({\bf x})\big\rangle\big\langle \mathscr{O}_i({\bf x}^\prime)\big\rangle\tag{2}$$ where $\mathscr{O}_i$ is the $i^{\rm th}$ component of a $n$-component order parameter ${\bf \mathscr{O}}$ and $\langle...\rangle$ denote thermal averages.
$1.$ If $G_{ij}({\bf x},{\bf x}^\prime)$ is identically zero, that implies $\mathscr{O}_i({\bf x})$ and $\mathscr{O}_i({\bf x}^\prime)$ are completely uncorrletaed or independent random variables, and vice-versa. 
$2.$ If the function $G_{ij}({\bf x},{\bf x}^\prime)$ decays exponentially with increasing $|{\bf x}-{\bf x}^\prime|$ and thus quickly vanishes beyond a certain length-scale $\xi(T)$, that implies there is a correlation but on small length-scales, and that presumably can be referred to as a short-range order. 
$3.$ At the phase transition point, $T=T_C$, where the function $G_{ij}({\bf x},{\bf x}^\prime)$ decays not exponentially but as some negative power of $|{\bf x}-{\bf x}^\prime|$, then that implies the divergence of the correlation length, i.e., $\xi(T_C)\to\infty$. At this point, perhaps, one would interpret this as the full system is trying to be correlated (?)
$4.$ If $G_{ij}({\bf x},{\bf x}^\prime)$ either remains constants or decays in such a way that even when $|{\bf x}-{\bf x}^\prime|\to\infty$, $G_{ij}({\bf x},{\bf x}^\prime)$ remains nonzero, finite, then there is the long-range order.

Question Apart from the four listed above, there is still one more important possibility. It is possible that the correlation function $G_{ij}({\bf x},{\bf x}^\prime)$ itself diverges at any temperature $T>0$. This is often the case for $2$-dimensional systems, for Hamiltonians with continuous symmetries and short-range interactions. This is at the heart of the Mermin-Wagner theorem. 
How do we properly interpret the meaning of this divergence) preferably from the defining equations in Eq.$(1)$ or Eq.$(2)$). In particular, why does a divergent correlation function should mean a lack of order in the system?
 A: Note that the 2-point function cannot diverge in models with bounded spins. So its divergence is clearly not at the heart of the Mermin-Wagner theorem. In fact, when the Mermin-Wagner theorem applies, the 2-point function usually still tends to 0 as the distance between the spins diverges. (See Chapter 9 of this book for more on this.)
In fact, even with unbounded spins, things usually cannot go wrong (when the infinite system is well defined!). Consider a real-valued field $(\varphi_x)$. If your field is translation invariant and the spins have a finite variance $\sigma^2$, then the Cauchy-Schwarz inequality implies
\begin{align}
\lvert\langle\varphi_0\varphi_x\rangle - \langle\varphi_0\rangle \langle\varphi_x\rangle\rvert
&=
\lvert\bigl\langle (\varphi_0-\langle\varphi_0\rangle)(\varphi_x-\langle\varphi_x\rangle)\bigr\rangle\rvert \\
&\leq
\sqrt{\bigl\langle(\varphi_0-\langle\varphi_0\rangle)^2\bigr\rangle \bigl\langle(\varphi_x-\langle\varphi_x\rangle)^2\bigr\rangle} \\
&=
\sigma^2,
\end{align}
which shows that the two-point function again cannot diverge.
So, you would either need to consider non-translation invariant models or models in which the spins have an infinite variance.
As a simple example of a non-translation-invariant model with diverging 2-point function, you can consider the massless Gaussian free field on $\mathbb{Z}^2$, with boundary condition $\varphi_0=0$. Note that you need to fix the value of one spin, or do something similar, since otherwise the infinite-volume field does not exist (as a consequence of the Mermin-Wagner theorem, see Section 9.3 of the same book for more on this).

Update: I collect here the content of the comments made by Abdelmalek Abdesselam or myself, in case they disappear.


*

*Your definition of long-range order (point 4 in your list) is not the standard one. Indeed, you should rather use the non-truncated 2-point function for this, since the truncated one always tends to zero as $|x|\to\infty$ in pure states.

*Your question seems to find its roots in some confusions regarding the divergences in some versions of the physicists' "proof" of the Mermin-Wagner theorem.
First, in many versions of the argument, the computation is carried by expressing the 2-point function $\langle S_0 \cdot S_x\rangle$ of the spin system (say an XY model) by a computation involving the two-dimensional massless Gaussian free field (after having approximated the cosine by a quadratic term and replaced the angle variables by real numbers). The correlation function $\bigl\langle(\varphi_0 - \varphi_x)^2\bigr\rangle$ does diverge (logarithmically) as $|x|\to\infty$ (this is closely related to the example I discuss above). This does not imply the divergence of the original 2-point function, which is necessarily finite, the spins being bounded. In fact, the 2-point function $\langle S_0 \cdot S_x\rangle$ actually tends to $0$ (as a power law) as $|x|\to\infty$.
Second, in some versions of the argument, there is a second divergence of the correlation in the GFF as the distance between the point tends to $0$. This is (1) due to the (totally useless) replacement of the lattice GFF by a continuum GFF and has no physical relevance in the context of the Mermin-Wagner theorem.
