Are the axioms for the real numbers consistent? Well, we know that for any set of axioms, it is necessary that no axiom or a conjunction of axioms contradict some other axiom. How do we know that this holds in the field, order and completeness axioms of the real number system?
 A: A general fact/insight in the foundations of mathematics is that one does not prove the consistency of a theory $T$ "out of thin air". If we want to prove that a theory $T$ is consistent (say, the set of axioms for a complete ordered field), we have to carry out that proof in some other axiom system, $S$. 
In the case of the axioms for a complete ordered field, we can construct a model of those axioms in the theory $S$ of ZFC set theory, which is a standard system for carrying out mathematics. We can also prove in ZFC that if a set of axioms has a model, then it is syntactically consistent. That means there is no formula $\phi$ such that both $\phi$ and the negation of $\phi$ are provable in the theory.
In the case of the theory of complete ordered fields, we can prove the consistency in much weaker systems. In particular, we can prove the consistency in a theory known by the acronym $\mathsf{ACA}_0$. The nice thing about that is that there is a proof in a much weaker system that $\mathsf{ACA}_0$ is consistent if and only if Peano arithmetic is consistent. Peano arithmetic is a standard set of axioms for the natural numbers with induction. 
So, in the end, the consistency of the axioms of complete ordered fields is no more "unreliable" than the consistency of Peano arithmetic. 
A: The reason that we know that is that the real numbers are a model of their axioms. And if a set of axioms has a model then it cannot prove a contradiction.
The reason is that if $T$ is any theory, and $T\vdash\varphi$, then in any model of $T$ we will have that $\varphi$ is true. This property is called soundness.
If $T$ is a first-order theory then the converse is true as well, if $\varphi$ is true in every model of $T$, then $T\vdash\varphi$. However note that the completeness axiom is not a first-order axiom (working in the language of fields), we quantify over all subsets of the real numbers.
But luckily, second-order logic is still sound, so if the axioms of the real numbers were inconsistent, it would prove a contradiction of the form $\varphi\land\lnot\varphi$. However in that case the real numbers themselves would satisfy both $\varphi$ and its negation. That is impossible, because of the way we define truth value for formulas in a structure guarantees that a structure cannot satisfy a statement and its negation.
