elementary number theory - The contradiction method used to prove that the square root of a prime is irrational

The contradiction method given in certain books to prove that sqare root of a prime is irrational also shows that sqare root of $4$ is irrational, so how is it acceptable?
e.g. Suppose $\sqrt{4}$ is rational,
$$\begin{align}
\sqrt{4} &=p/q \qquad\text{where pand q are coprimes} \\
4 &=p^2/q^2\\
4q^2&=p^2 \tag{1} \\

4&\mid p^2\\
4&\mid p\\
\text {let }p&=4m \qquad\text{for some natural no. m} \\
p^2&=16m^2\\
4q^2&=16m^2 \qquad\text{(from (1) )}\\
q^2&=4m^2\\
4& \mid q^2\\
4&\mid q
\end{align}
$$

but this contradicts our assumption that $p$ and $q$ are coprime since they have a common factor $p$. Hence $\sqrt{4}$ is not rational. But we know that it is a rational. Why?