Argument on why spin correlation functions in Ising model decay exponentially with a correlation length?

Let me write the Hamiltonian $$ H = -J sum_i S_i^z S_{i+1}^z. $$ This choice will avoid some annoying (and irrelevant) signs.

One way to formulate the statement in the OP precisely is as follows.

Consider the variables $delta_i=S_i^zS_{i+1}^z$. Since $delta_i=1$ when the spins at $i$ and $i+1$ agree and $delta_i=-1$ when the spins at $i$ and $i+1$ disagree, you can identify them with the kinks in your question (that is, there is a kink between $i$ and $i+1$ when $delta_i=-1$).

Introducing the variables $delta_i=S_i^zS_{i+1}^z$, the Hamiltonian becomes $$ H = J^z sum_i delta_i. $$ It follows that the random variables $delta_i$ are independent and identically distributed. One can easily compute their expectation: since $$ P(delta_i = 1) = frac{e^{beta J^z}}{e^{beta J^z} + e^{-beta J^z}}, $$ one has $$ langle delta_i rangle = frac{e^{beta J^z} - e^{-beta J^z}}{e^{beta J^z} + e^{-beta J^z}} = tanh(beta J^z). $$ Finally, noting that $S_i^zS_{i+r}^z = delta_idelta_{i+1}cdotsdelta_{i+r-1}$, we obtain $$ langle{S_i^zS_{i+r}^z}rangle = langledelta_idelta_{i+1}cdotsdelta_{i+r-1}rangle = langle delta_i rangle^r = (tanh(beta J^z))^r. $$

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In words, the fact that kinks proliferate in the system (at each $i$, there is a positive probability that a kink is present, so there will be a positive density of them in the system) prevents the ordering of the spins.