Is it possible to prove a mathematical statement by proving that a proof exists? I'm sure there are easy ways of proving things using, well... any other method besides this!
But still, I'm curious to know whether it would be acceptable/if it has been done before?
 A: There is a disappointing way of answering your question affirmatively: If $\phi$ is a statement such that First order Peano Arithmetic $\mathsf{PA}$ proves "$\phi$ is provable", then in fact $\mathsf{PA}$ also proves $\phi$. You can replace here $\mathsf{PA}$ with $\mathsf{ZF}$ (Zermelo Fraenkel set theory) or your usual or favorite first order formalization of mathematics. In a sense, this is exactly what you were asking: If we can prove that there is a proof, then there is a proof. On the other hand, this is actually unsatisfactory because there are no known natural examples of statements $\phi$ for which it is actually easier to prove that there is a proof rather than actually finding it. 
(The above has a neat formal counterpart, Löb's theorem, that states that if $\mathsf{PA}$ can prove "If $\phi$ is provable, then $\phi$", then in fact $\mathsf{PA}$ can prove $\phi$.)
There are other ways of answering affirmatively your question. For example, it is a theorem of $\mathsf{ZF}$ that if $\phi$ is a $\Pi^0_1$ statement and $\mathsf{PA}$ does not prove its negation, then $\phi$ is true. To be $\Pi^0_1$ means that $\phi$ is of the form "For all natural numbers $n$, $R(n)$", where $R$ is a recursive statement (that is, there is an algorithm that, for each input $n$,  returns in a finite amount of time whether $R(n)$ is true or false). Many natural and interesting statements are $\Pi^0_1$: The Riemann hypothesis, the Goldbach conjecture, etc. It would be fantastic to verify some such $\phi$ this way. On the other hand, there is no scenario for achieving anything like this.
The key to the results above is that $\mathsf{PA}$, and $\mathsf{ZF}$, and any reasonable formalization of mathematics, are arithmetically sound, meaning that their theorems about natural numbers are actually true in the standard model of arithmetic. The first paragraph is a consequence of arithmetic soundness. The third paragraph is a consequence of the fact that $\mathsf{PA}$ proves all true $\Sigma^0_1$-statements. (Much less than $\mathsf{PA}$ suffices here, usually one refers to Robinson's arithmetic $Q$.) I do not recall whether this property has a standard name. 
Here are two related posts on MO: 


*

*$\mathrm{Provable}(P)\Rightarrow \mathrm{provable}(\mathrm{provable}(P))$?

*When does $ZFC \vdash\ ' ZFC \vdash \varphi\ '$ imply $ZFC \vdash \varphi$?
A: I'd say the model-theoretic proof of the Ax-Grothendieck theorem falls into this category.  There may be other ways of proving it, but this is the only proof I saw in grad school, and it's pretty natural if you know model theory.
The theorem states that for any polynomial map $f:\mathbb{C}^n \to\mathbb{C}^n$, if $f$ is injective (one-to-one), then it is surjective (onto).  The theorem uses several results in model theory, and the argument goes roughly as follows. 
Let $ACL_p$ denote the theory of algebraically closed fields of characteristic $p$.  $ACL_0$ is axiomatized by the axioms of an algebraically closed field and the axiom scheme $\psi_2, \psi_3, \psi_4,\ldots$, where $\psi_k$ is the statement "for all $x \neq 0$, $k x \neq 0$".  Note that all $\psi_k$ are also proved by $ACL_p$, if $p$ does not divide $k$.


*

*The theorem is true in $ACL_p$, $p>0$.  This can be easily shown by contradiction: assume a counter example, then take the finite field generated by the elements in the counter-example, call that finite field $F_0$.  Since $F_0^n\subseteq F^n$ is finite, and the map is injective, it must be surjective as well.

*The theory of algebraically closed fields in characteristic $p$ is complete (i.e. the standard axioms prove or disprove all statements expressible in the first order language of rings).

*For each degree $d$ and dimension $n$, restrict Ax-Grothendieck to a statement $\phi_{d,n}$, which is expressible as a statement in the first order language of rings. Then $\phi_{d,n}$ is provable in $ACL_p$ for all characteristics $p > 0$.

*Assume the $\phi_{d,n}$ is false for $p=0$.  Then by completeness, there is a proof $P$ of $\neg \phi_{d,n}$ in $ALC_0$.  By the finiteness of proofs, there exists a finite subset of axioms for $ACL_0$ which are used in this proof.  If none of the $\psi_k$ are used $P$, then $\neg \phi_{d,n}$ is true of all algebraically closed fields, which cannot be the case by (2).  Let $k_0,\ldots, k_m$ be the collection of indices of $\psi_k$ used in $P$.  Pick a prime $p_0$ which does not divide any of $k_0,\ldots,k_m$.  Then all of the axioms used in $P$ are also true of $ACL_{p_0}$.  Thus $ACL_{p_0}$ also proves $\neg \phi_{d,n}$, also contradicting (2).  Contradiction.  Therefore there is a proof of $\phi_{d,n}$ in $ACL_0$.


So the proof is actually along the lines of "for each degree $d$ and dimension $n$ there is a proof of the Ax-Grothendieck theorem restricted to that degree and dimension."  What any of those proofs are, I have no clue.
