Why is this definition of convergence incorrect?

I am not sure why the other current answers are interpreting your second definition as invalid, because as a native English speaker I interpret it to mean exactly the same (logically speaking) as the first definition. In particular, you did not switch the quantifiers, even though the surface text appears to have them switched. You wrote:

There exists an index $N$ for every $\epsilon > 0$, such that for all $n \geq N$, we have ...

This means that there is some $N$ for every $\epsilon > 0$, not necessarily the same $N$ for all $\epsilon > 0$. So it conveys the same logical structure as the other definition, whereby for any given $\epsilon > 0$ there is an index $N$ such that ...

It would become invalid if you had the following English phrasing (which switches the quantifiers):

There exists an index $N$ such that, for every $\epsilon > 0$ and for all $n \geq N$, we have ...

But you did not use that phrasing, so that does not imply anything about yours.

Just for fun, here is a quote (apocryphally attributed to Albert Einstein) using exactly that phrase structure:

Stay away from negative people. They have a problem for every solution.

It is clearly understood by everyone to mean:

Stay away from negative people, because for every solution they will find some problem with it.

I think, based on your wording and the answers / comments, that there is a misunderstanding in the way things are written / thought.

There is a difference between

$\exists N$ such that $\forall \epsilon>0$, such that $\forall n>N$, $|a_{n} - a| < \epsilon$

and

$\exists N_\epsilon$ for all $\epsilon>0$, such that $\forall n>N_\epsilon$, $|a_{n} - a| < \epsilon$

You wrote the first (it is at least implied), hence the answers you got. I believe you were meaning the second. The second is, roughly, equivalent to the initial definition you are mentionning. But it is a bit of an awkward formulation, open to misunderstanding (as you can see), even with proper syntax.

Look at your second definition and take a $N$ such that for every $\varepsilon > 0$, $\forall n \geq N$, $|a_n-a| \leq \varepsilon$.

This means that for all $n \geq N$, $|a_n - a|$ is as small as you want. So it has to be zero. So your definition is equivalent to $$\exists N \in \mathbb{N}, \forall n\geq N, a_n = a$$

i.e. $(a_n)$ is stationary. Of course this is not equivalent to the convergence.