In College, I was taught that the CI for proportion is as follows.
$$ \left( \ \overline{\!X}\:\pm \frac{1}{2\sqrt{n\cdot \alpha }} \right) $$
But this is very far from the one that is commonly known, that is
$$ \hat{p}\:\pm z * \ \sqrt\frac{p(1-p)}{n} $$
Can anyone please elaborate where does the first CI come from?
In your first expression, I am not sure what $\alpha$ is or why the $2$ remains in the denominator
In your second expression, I think you are missing a square root so might be looking at $$\hat{p}\:\pm z \sqrt{ \frac{p(1-p)}{n}}$$
The second part of that is maximised when $p=\frac12$ making $p(1-p)=\frac14$ and $\sqrt{ \frac{p(1-p)}{n}} =\frac1{2\sqrt{n}}$
If $z \approx 1.96$ based on a $95\%$ two-tailed confidence interval for a normal distribution and you round this to $2$, and you use the natural estimator $\hat p=\bar X$, you would get $$\bar X \pm \frac1{\sqrt{n}}$$
This may be unnecessarily wide when $p$ is not close to $\frac12$ but at least is conservative. Depending on your point of view, it is simple or crude and there may be more sophisticated methods for binomial proportion confidence intervals when $n$ is small