Two-point functions don't signal spontaneous symmetry breaking? I am aware that when some operator $\phi_n(x)$ which transforms nontrivially with respect to some symmetry group acquires a VEV, it signals the spontaneous breakdown of a particular symmetry, since there must be different vacua to accomodate for the different values of $\langle\phi_n(x)\rangle$ under the symmetry transformation.
My question is, why doesn't the same logic apply to two-point functions like $\langle T\phi_n(x)\phi_m(y)\rangle$? These are expected to be nonzero, and they transform nontrivially. Somehow we don't make the conclusion that a nonzero two-point function implies spontaneous symmetry breaking. 
What I am thinking:
I can make an analogy to a different situation that might help the discussion. If we consider Yang-Mills theory, there are 1-form global symmetries, which act on Wilson and 't Hooft line operators. This 1-form symmetry can be spontaneously broken, and the criterion is as follows. If a Wilson loop $\langle W(L)\rangle$ follows an area law for large loops the symmetry is unbroken, while if it follows a circumference law the symmetry is broken. The logic for this is that a circumference law can be canceled by some local counterterm, and so the Wilson loop won't go to zero for large loops.
The situation may be similar for the two-point case, where it can either have an exponential decay (which signals a mass gap) or some slower decay. 
However, if this is the case it seems weird to me that we don't care what happens when $x\to y$ (which probes the UV), we only care about about the IR behavior of the two point function.
Response to the singlet answers
The field strength two point correlation function is 
$$\langle T F_{\mu\nu}(x)F_{\lambda\sigma}(0)\rangle=\frac{4}{(x^2)^2}\Big(\eta_{\mu\lambda}\eta_{\nu\sigma}-2\eta_{\nu\sigma}\frac{x_{\mu}x_{\lambda}}{x^2}-2\eta_{\mu\lambda}\frac{x_{\nu}x_{\sigma}}{x^2}-(\mu\leftrightarrow\nu)\Big)$$
Looking at the tensor structure of this, the first term with just $\eta$'s is certainly a singlet of the Lorentz group, but I don't see how the other parts are singlets. This suggests that Lorentz symmetry is spontaneously broken based on what you are saying. 
Response to MannyC's answer
While I agree that what you did shows no contradiction with the fact that $Q|\Omega\rangle=0$ and the correlator is non-zero, there does appear to be a contradiction by considering the quantity 
$$0\neq\langle\Omega|[Q,\phi(x_1)O_1(x_2)\cdots O_{n-1}(x_n)]|\Omega\rangle=0$$
This is nonzero by the assumption that the operator transforms non-trivially, but it is zero by the assumption that $Q|\Omega\rangle=0$. So actually, if $Q|\Omega\rangle=0$ then the correlator must be in the singlet representation. This observation doesn't bode well for the two point correlator of field strengths however. 
 A: In a field theory with O(N) symmetry, spontaneous symmetry breaking implies that $\varphi_n\equiv\langle \phi_n\rangle\neq0$. The two-point correlators can then be written in all generality as
$$
\langle\phi_n(x)\phi_m(y)\rangle=\delta_{nm}G_1(x,y)+\varphi_n\varphi_m G_2(x,y),
$$ 
where $G_1$ and $G_2$ are invariant functions (which will in general depend on $\sum_n\varphi_n^2$. Thus, a non-trivial tensor structure of the correlation function implies spontaneous symmetry breaking.
On the other hand, the absence of spontaneous symmetry breaking just implies that $\varphi_n=0$ and thus $\langle\phi_n(x)\phi_m(y)\rangle=\delta_{nm}G_1(x,y)$, which transforms trivially under $O(N)$.
A: I have two different ways to look at it.
Charge acting on the vacuum
Take a symmetry group $G$ and a field $\phi$ which is not a singlet. There is a Cartan generator $Q$ in $G$ that is non vanishing on $\phi$, namely
$$
[Q, \phi ] = q_\phi \phi\,, 
$$
for some non zero number $q_\phi$. Let me denote the vacuum as $|\Omega\rangle$. If $\phi$ has a non zero one-point function then
$$
0 \neq q_\phi\langle \Omega |\phi| \Omega\rangle = \langle \Omega| [Q,\phi]|\Omega\rangle\,.
$$
If this has to be non-zero, then necessarily $Q |\Omega\rangle \neq 0$, thus the vacuum is not invariant. On the other hand, any other $n$-point funtion will be
$$
q_\phi \langle \Omega|\phi(x_1)\,O_1(x_2)\cdots O_{n-1}(x_n)|\Omega\rangle = \langle \Omega| [Q,\phi(x_1)]\,O_1(x_2)\cdots O_{n-1}(x_n)|\Omega\rangle\,.
$$
Even if $\langle \Omega | Q = 0$ there will be a piece like
$$
\langle \Omega| \phi(x_1)\,Q\,O_1(x_2)\cdots O_{n-1}(x_n)|\Omega\rangle\,,
$$
which doesn't have to vanish. So a non vanishing $(n>1)$-point function is consistent with a $G$ invariant vacuum.

Wigner-Eckart-like argument
Another point of view, which came up in the comments, is to think of it in the same way as the Wigner-Eckart theorem. Take two operators $O_1$, $O_2$ in the representations $\rho_1$ and $\rho_2$ of some group $G$. If the vacuum is $G$ invariant (i.e. no spontaneous symmetry breaking), the expectation value
$$
\langle \Omega |O_1(x_1) O_2(x_2) | \Omega\rangle\,,
$$
will be proportional to some matrix element
$$
\langle \Omega | T_{12} |\Omega \rangle \,,
$$
where $T_{12}$ is kind of a "placeholder" operator that transforms under the representations in the tensor product $\rho_1 \otimes \rho_2$. If such a tensor product contains a singlet, the above matrix element does not need to vanish. Clearly even when $\rho_1$ and $\rho_2$ are non trivial one can find singlets in $\rho_1\otimes \rho_2$.
