Consider a product space $X = \{0,1\}^\mathbb{Z}$ and the space of probability measures on $X$, $\mathcal{M}(X)$. We say that for any two $a, b \in X$, $$a \prec b \iff a_x \leq b_x \, \, \, \, \, \forall x \in \mathbb{Z},$$ where the subscript denotes the $x$-th component of the vector $a$. We say that the function $f : X \rightarrow X$ is$$ \mbox{ monotone }\iff \forall a,b \in X \mbox{ s. t. } a \prec b, \, \, \, \, \, f(a) \leq f(b).$$ We say that for any $\mu, \nu \in \mathcal{M}(X)$, $$\mu \prec \nu \iff \forall f \mbox{ monotone } , \, \, \, \int f d\mu \leq \int f d \nu.$$
Term a set $C \subset X$ full-below if $$\forall b \in C,\, \, a \prec b \implies a \in C.$$
Question: How to prove the following statement? $$ \mu( C) \geq \nu( C) \, \, \, \, \forall C \mbox { full below } \implies \mu \prec \nu$$
I think there is a typo in the definition of the order on $X$, I think you mean $\forall x \in \mathbb Z$?
I'm going to assume that you consider monotone functions $f: X \to \mathbb{R}_{\geq 0}$ such that $f$ is measurable. Potentially you could get rid of the measurability assumption by assuming that the $\sigma$-algebra contains all sets full-below(equivalently full-above) and then show that all monotone functions are measurable.
I define $$C \subset X \text{ full-above } \Longleftrightarrow \forall x \in C: x \prec y \Rightarrow y \in C.$$ We observe that $C$ is full-above iff $C^c$ is full-below. For let $x \in C^c$ assume that $y \prec x$ and $y \in C$ this would imply $x \in C$ a contradiction.
Now to prove the statement, we will as usual start by proving it for simple functions, so we first need to identify the simple monotone functions. Assume that we have a simple function $f$ that only takes 2 values $0 = a_0$ and $a_1 >0$, then we must have $f = a_1 1_{C}$ i.e. the indicator function on some set $C$. Assume $x \in C$ and $x \prec y$ then $a_1 = f(x) \leq f(y)$ so we must have $f(y) = a_1$, hence $C$ is full-above. But then $C^c$ is full-below and we must have $$1-\mu(C) = \mu(C^c) \geq \nu(C^c) = 1 - \nu(C).$$ This implies that $\mu(C) \leq \nu(C)$ and hence $\int f d\mu \leq \int f d\nu$(remember our coefficients are non-negative).
Assume that we have a simple, monotone function $f = \sum_{i=0}^n a_i 1_{C_i}$ with $0 = a_0 < a_1 < \dots < a_n$ and the $C_i$ disjoint. As before we see that $C_n$ must be full-above. Similarly we can conclude that $D_m \cup_{i=m}^n C_i$ is full-above for all $m$. But this gives us $$ a_1\mu(D_1) \leq a_1 \nu(D_1).$$ But we also see that $$ (a_2 - a_1) \mu(D_2) \leq (a_2 - a_1) \nu(D_2) $$ adding the two in equalities give $$ a_1\mu(C_1) + a_2 \mu(D_2) \leq a_1 \nu(C_1) + a_2 \mu(D_2). $$ In general we see that $$ (a_m - a_{m-1})\mu(D_m) \leq (a_m - a_{m-1}) \nu(D_m) $$ hence by induction $$ a_1\mu(C_1) + \dots + a_{m-1}\mu(C_{m-1}) + a_m \mu(D_m) \leq a_1 \nu(C_1) + \dots + a_{m-1}\nu(C_{m-1}) + a_m \nu(C_m).$$ But this proves $$ \int f d\mu \leq \int f d\nu.$$
Now the only thing left is to realise that we can approximate any monotone function by simple monotone functions. But observe that if $f$ is monotone then $f^{-1}([a, \infty))$ is full-above, similarly $f^{-1}((a, \infty))$ is full-above. Hence for $m > 0$ put $C_i = f^{-1}([\frac{i}{2^m}, \frac{i+1}{2^m}))$ and $a_i = \frac{i}{2^m}$ for $i \leq n = m*2^m$ then if we set $$ g_m = \sum_{i=0}^n a_i 1_{C_i}. $$ We must have $g_{m} \leq g_{m'}$ for $m \leq m'$ and $\lim g_m(x) = f(x)$ i.e. we have monotone, point-wise convergence of positive functions. Furthermore we see that the $g_m$ are monotone since $D_i = f^{-1}([\frac{i}{2^m}, \infty))$ is full-above for all $i$. Thus we get $$ \int f d\mu = \lim_{m \to \infty} \int g_m d\mu \leq \lim_{m \to \infty} \int g_m d\nu = \int f d\nu. $$