A silly question with unprovability By Gödel's incompleteness theorems, we can get a true but unprovable sentance $\psi$. However, we know it is true since its falsity implies contradiction. Then, why couldn't we accept this "proof" since it shows that $\psi$ must be true? Does it mean that the law of excluded middle could not be written as an axiom of finite length?
 A: We start with an appropriate theory $T$ and make the Gödel sentence $G_T$. When we argue that $G_T$ is true, we do not make that argument within $T$. We generally need to assume something beyond $T$ - such as "$T$ is consistent" - in order to show that $G_T$ is true.  But, if $T$ is an appropriate theory, the incompleteness theorems show that $T$ does not prove "$T$ is consistent" and so this argument can't work within $T$. 
So the real issue with the phrase in the question is not with the word "true", the issue is with the word "unprovable". In the motto "true but unprovable", the term "true" refers to the standard model, while the phrase "unprovable" only means "unprovable in $T$", not "unprovable in any system whatsoever".  Every statement is provable in some system, such as a system that includes the statement as an axiom.  
A: In Godel's incompleteness theorem, completely decidable (strongly representable) predicate $P$ is created with the properties:
$$\forall n (\vdash P(n))$$
$$\not \vdash \forall n ~P(n)$$
This $P$ is not the Godel sentence $G$, neither is $\forall n~P(n)$.  The Godel sentence is the statement made by a diagonalization argument based on $P$.
So when the claim is made that $\forall n~P(n)$ is true, it means true in the demonstrable sense.  That means there is a proof for $P(0)$, and for $P(1)$, etc.  Any number you pick, there is a proof. But when it is said $\forall n ~ P(n)$ is unprovable, they mean that despite there being a proof for each instance of $P$, no matter how cleverly you picked the rules of inference concerning $\forall$ to be, once the entire logic is known, such a $\forall n~ P(n)$ can be constructed that has no proof.

Then, why couldn't we accept this "proof" since it shows that $\psi$ must be true?

There are 2 logics that go into Godels incompleteness theorem.  There is the logic that the theorem is referring to, let's call it $L_1$, and there is the logic or generally reasoning used to construct the proof of the claim about $L_1$, call it $L_2$.  It is common for $L_2$ to be just model theoretic reasoning, but it can also be completely reliable constructive algorithmic reasoning, since the proof itself is of an existential claim (that such a $P$ can be constructed).
When it is said that we know $\forall n \vdash P(n)$, that is a construction done in $L_2$, not $L_1$.  So when you say "couldn't we accept this proof", the answer is yes we can, and yes we do, but we do so in the logic of $L_2$.  But the entire point is that the statement isn't provable in $L_1$.  So you might ask "but then doesn't Godel's incompleteness theorem not apply to $L_2$?" and the answer would be "it doesn't apply to $L_2$ for that choice of $P$, but it does apply for another choice of $P$, which could then be proven in a more presumptuous logic $L_3$".
