Prove that number $\sqrt{2}$ is an irrational number using this theorem “if $a^2$ is even, $a$ must be even”

Question

I have to prove the following:

Prove $\sqrt{2}$ is an irrational number using this theorem "if $a^2$ is even, $a$ must be even"

I made a proof by contradiction for the statement above, but I believed that there is a mistake in my proof.
Suppose $\sqrt{2}$ is a rational number,

let a = $\sqrt{2}$
\begin{align}
a = \sqrt{2}\\
a^2 = 2
\end{align}
Therefore, $a^2$ is even
Given that if $a^2$ is even, $a$ must be even,

but $\sqrt{2}$ is not even
Hence, we have a contradiction that if $a^2$ is even, $a$ must be even
Thus, $\sqrt{2}$ is an irrational number

Question: Would someone mind pointing out where my mistake is and why it is a mistake?
Thank you for your kind attention

Answer 1

I prove the statement so you can see how this kind of proofs are done:

For the sake of contradiction, suppose that $\sqrt{2}$ is rational. We can write $$\sqrt{2}=\frac{p}{q},$$ where $p,q$ are integers and the RHS fraction is irreducible. Squaring both sides and multiplying by $q^2$ we get $$2q^2=p^2.$$ Since, $2q^2$ is even then $p^2$ is even. Can you follow from here?