Here is the question:
A surveyor conducts research on 4148 households, randomly selected. An estimate of $7.9\%$ unemployment was obtained from a survey of $4148$ people.
Construct a $95\%$ confidence interval with this data.
Solution:
An approximate $95\%$ confidence interval for the true unemployment rate $p$ is $$\left(\hat{p} - z_{2.5}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}},\hat{p} - z_{97.5}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right)\\=\left(0.079 - 1.96\sqrt{\frac{0.079(1-0.079)}{4148}},0.079 + 1.96\sqrt{\frac{0.079(1-0.079)}{4148}}\right)\\=(0.071,0.087)$$
What I'm mainly confused about is: how did they jump immediately to their first line?
I feel as if they made the assumption that each person is bernoulli distributed? And then they assumed that when you standardize the average, this is normally distributed?
Are these assumptions valid to make?
Note the small typo in the first line of your formula ( the upper limit of the interval should be $+$)
It's a standard result that an approximate $100(1-\alpha)\%$ CI for a proportion $p$, obtained by observing $x$ successes in a sequence of $n$ independent Bernoulli trials each with success probability $p$ is $$ \left(\hat{p} - z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}},\hat{p} + z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right) $$ where $\hat{p}=x/n$ is the estimate of $p$ and $z$ is the $(1-\alpha/2)$ quantile of the standard normal distribution.
So yes, assumptions have been made about independent Bernoulli trials.