"How can Peano Arithmetic ever be proved consistent?"
I assume you're asking for an absolute proof of consistency of (first-order) PA. That was more or less Hilbert's goal, namely to give a finitist proof of consistency of mathematics (yet to be defined but including arithmetic). The incompleteness theorems totally and finitistically demolished any hope of that, and any philosophers who think otherwise do not actually grasp the mathematics.
Furthermore, any practical foundational system whatsoever must necessarily be a syntactic system, and the mere acceptance of any such system requires ontological commitment to the basic properties of finite strings of symbols (whether abstract or in any physical medium), including:
- Closure of strings under concatenation $+$.
- Existence of an empty string $e$, namely $e+x = x = x+e$ for any string $x$.
- Associativity of $+$ on strings, namely $(x+y)+z = x+(y+z)$ for any strings $x,y,z$.
- Existence of distinct symbols $p,q$, namely distinct non-empty strings $p,q$ such that
$p \ne u+v$ and $q \ne u+v$ for any non-empty strings $u,v$.
- Given any strings $a,b,c,d$ such that $a+b = c+d$, there is a string $x$ such that
either ( $a+x = c$ and $b = x+d$ ) or ( $a = c+x$ and $x+b = d$ ).
It turns out that these five properties alone are sufficient to give rise to what is called the theory of concatenation, which is essentially incomplete but can already express very natural translations of statements about the halting behaviour of coded Turing machines on coded inputs. Therefore, the acceptance of meaningfulness of syntactic systems already commits you to accept something that already is strong enough to do basic arithmetic and yet is essentially incomplete. (Note that the phrase "essentially incomplete" is a technical term, not linguistic emphasis!)
But you also made numerous fundamental mistakes:
"Peano Arithmetic cannot prove itself consistent by the incompleteness theorem."
PA can prove the arithmetic statement Con(PA) if PA is itself inconsistent. Since you do not know whether PA is consistent, you cannot a priori exclude absolutely the possibility that PA proves itself consistent in this sense.
"So it requires some higher order theory."
Wrong again. PA+Con(PA) is a first-order theory that trivially proves the consistency of PA. In fact, one may not even need a strictly stronger system to prove the consistency of PA. For instance, just the theory with a single axiom Con(PA) would also do the job. Does it have meaning? No. But it serves to refute the claim that you need a stronger theory.
"That may prove Peano [Arithmetic] consistent, but that proof itself can only be relied upon if the higher order theorem is itself consistent. Which it in turn cannot prove itself."
Generously interpreted, this is vague but true, and is much more precisely stated as the essential incompleteness of PA (or just Robinson's Q). I have already mentioned to you a number of times before that you can see a complete proof of this fact here.
"It would appear therefore, by induction that Peano Arithmetic can never be proven consistent."
False. As proven in my linked post, there is no consistent useful formal system that interprets PA and proves itself consistent, where "useful" means "has a proof verifier" as defined in that post. This has nothing to do with induction. We can never prove that PA cannot be proven consistent, simply because we can easily do so in an inconsistent system, which for all we know PA itself might be.
Curiously though, PA itself is strong enough to prove Con(PA$_k$) where PA$_k$ is the fragment of PA that is PA$^-$ plus induction for $1$-parameter sentences with at most $k$ quantifiers, for every natural $k$ (in the meta-system). PA is simply unable to prove the universal sentence $\forall k\ ( \text{Con}(\text{PA}_k) )$ although it can prove every instantiation of it on a natural, so this is an example of the $ω$-incompleteness of PA that is simultaneously striking as a 'near-miss' of self-verified consistency.
