Why the fact that $\mathbb{N}$ has decidable equality does not imply $\mathsf{HA} = \mathsf{PA}$? It is well-known that the natural numbers has decidable equality, the property that for any $n, m \in \mathbb{N}$
$$ n=m \lor n \neq m .$$
In fact, this is not so difficult to prove by double induction. Moreover, equality is the only relation contained in the signature of intuitionistic number theory ($\mathsf{HA}$), meaning that any atomic proposition of it will have the form of $n = m$.
So why is that this does not imply the law of excluded middle, which says that
$$ P \lor \neg P$$
for every proposition $P$? Clearly, we have the base case. Does this argument breaks down in the inductive step? In this case, it fails for which connectives?  
Disjunction, perhaps? Well, it doesn't seem so. For suppose that $P$ has the form $ (n = m \lor i = j)$. Then our goal
$$ (n = m \lor i = j) \lor \neg (n = m \lor i = j) $$
is the same as
$$ (n = m \lor i = j) \lor (n \neq m \land i \neq j) $$
which by distributivity we obtain
$$ (n \neq m \lor (n = m \lor i = j)) \land (i \neq j \lor (n = m \lor i = j))$$
and by associativity we have
$$ ((n = m \lor n \neq m) \lor i = j) \land (n = m \lor (i = j \lor i \neq j))$$
this, however, is true, since both $n=m \lor n \neq m$ and $i=j \lor i \neq j$ are the case. So where exactly is the point the argument breaks down? Otherwise, $\mathsf{HA}=\mathsf{PA}$, which is obviously false.

It's the induction step where this breaks down - specifically, once we introduce quantifiers. Consider a sentence of the form $\forall x\exists y\varphi(x, y)$ where there is no computable function $f$ such that for all naturals $n$, $\varphi(n, f(n))$ holds (coming up with such a sentence is a good exercise). LEM will fail for such a sentence in HA.
