The proof that $1-\sqrt{-5}$ is irreducible is incorrect. In addition to a minor (and alas, common) mistake in writing that makes what you write not what you intended to write, there is an assertion which is just plain false.
Explicilty, you write:
Assume $1-\sqrt{-5}$ is reducible, then there must exist $a,b \in \mathbb{Z}[\sqrt{-5}]$ so that $N(1-\sqrt{-5})=N(a)N(b) \Rightarrow N(a)=N(b)= \pm (1-\sqrt{-5})$
First: the use of $\Rightarrow$ is incorrect. What you have written is that there exist $a$ and $b$ in $\mathbb{Z}[\sqrt{-5}]$ for which the following statement holds:
If $N(1-\sqrt{-5}) = N(a)N(b)$, then $N(a)=N(b) = \pm (1-\sqrt{-5})$.
What you actually wanted to write was that
If there exist $a,b\in\mathbb{Z}[\sqrt{-5}]$ such that $N(1-\sqrt{-5}) = N(a)N(b)$, then $N(a)=N(b)=\pm(1-\sqrt{-5})$.
What's the difference? The first statement will be true if you can find an $a$ and a $b$ for which $N(1-\sqrt{-5})$ is not equal to $N(a)N(b)$! It will be true that the implication holds, because the antecedent will be false. So, exhibiting $a=7$ and $b=10578432$ makes the statement you wrote true. However, they are irrelevant towards the second statement (and towards establishing what you want to establish, namely, that no such $a$ and $b$ exist).
Second: this is incorrect. What you want to assume is that there exist $a$ and $b$ such that $1-\sqrt{-5} = ab$, and neither $a$ nor $b$ are units; you do not simply want to assume that the product of the norms of $a$ and $b$ equals the norm of $1-\sqrt{-5}$.
Third: Even so, you conclusion is nonsense. The norm of any element of $\mathbb{Z}[\sqrt{-5}]$ must be an integer. What you want to conclude is that $N(a)N(b) = N(1-\sqrt{-5}) = 6$, and then get a contradiction. The norm cannot equal $\pm(1-\sqrt{-5})$.
It's also nonsense to say "since $1-\sqrt{-5}$ is not a quadratic remainder modulo $5$". It's not even a remainder modulo $5$, because it's not an integer!
The argument about $2$ not being prime is likewise incorrect in its use of the norm, which quickly reduces to nonsense symbols being strewn around:
But with $a,b \in \mathbb{Z}[\sqrt{-5}]$ it immediately follows that for :
$(1\pm\sqrt{-5})= 2(a+b\sqrt{-5})$ $2b = \pm 1$. So 2 is not a prime in $\mathbb{Z}[\sqrt{-5}]$.
This is nonsensical as written. I suspect you wanted to say something like
But there cannot exist $a,b\in\mathbb{Z}$ (not in $\mathbb{Z}[\sqrt{-5}]$) such that $2(a+b\sqrt{-5}) = 2a+2b\sqrt{-5} = 1-\sqrt{-5}$.
Same issues with the rest of the arguments. They are either terse to the point of nonsense, or contain incorrect or incoherent claims. I strongly urge you to write complete sentences, use words, and don't over rely on symbols. And to read your own arguments with a critical eye after you are done.
Best Answer
To show that $2$ is not prime, observe that $2$ divides $(\sqrt{5}+1)^2$ but $2$ does not divide $\sqrt{5}+1$. Or else we can use $(\sqrt{5}-1)(\sqrt{5}+1)$.
Similarly, we can show that $\sqrt{5}+1$ is not prime. For $(\sqrt{5}+1)(\sqrt{5}-1)=4$. So $\sqrt{5}+1$ divides $(20(2)$, but $\sqrt{5}+1$ does not divide $2$.
To show that $a^2-5b^2$ cannot be equal to $\pm 2$, observe that any square is congruent to $0$, $1$, or $-1$ modulo $5$. Since $a^2-5b^2\equiv a^2\pmod{5}$, we cannot have $a^2-5b^2$ equal to anything congruent to $\pm 2\pmod{5}$. In particular, it cannot be equal to $2$ or $-2$.
Remark: It turns out that this flaw can be fixed. If we consider the numbers of the form $a+b\sqrt{5}$, where $a$ and $b$ are integers, together with numbers of the form $\frac{a+b\sqrt{5}}{2}$, where $a$ and $b$ are odd integers, we get a structure in which every non-unit irreducible is prime.