When constructing confidence intervals usually the size of a population is far larger than the sample size. In these cases we treat the sample as if it came from an infinite population and this simplifies the analysis a bit. For these cases the confidence interval formula is the following
Lower limit:
$$p-z\sqrt{\frac{p(1-p)}{n}}$$
For your example this is $0.5-1.96\sqrt{\frac{0.5(1-0.5)}{1406}}=0.4739$
Upper limit:
$$p+z\sqrt{\frac{p(1-p)}{n}}$$
For your example this is $0.5261$ so the 95% confidence interval for the population value of $p$ is $(0.4739,0.5261)$
Small population size
When the size of the population is small then you can make an adjustment to account for this fact. In this case the confidence interval is
Lower limit:
$$p-z\sqrt{\frac{p(1-p)}{n}\left(\frac{N-n}{N-1} \right)}$$
Upper limit:
$$p+z\sqrt{\frac{p(1-p)}{n}\left(\frac{N-n}{N-1} \right)}$$
The part under the square root is modified slightly. In your example the population is huge so it's being modified by a factor of $\frac{292456752-1406}{292456752-1}= 0.999995$. You can try calculating the modified confidence interval, it doesn't change the first four decimal places.
Small sample sizes
When you sample very few people then the methods used to derive the above formulas can be invalid. A common rule for deciding if sample size is large enough is the following:
If $np > 5$ and $n(1-p)>5$ then the sample size is large enough. Your example certainly has a large enough sample size. When the sample size is too small then you should use a different interval such as the Wilson Score interval:
$$\text{Lower limit} = \frac { 2n\hat{p} + z^2 - \left[z \sqrt{z^2 - \frac{1}{n} + 4n\hat{p}(1 - \hat{p}) + (4\hat{p} - 2)} + 1\right] }
{ 2(n + z^2) }$$
$$\text{Upper limit} = \frac { 2n\hat{p} + z^2 + \left[z \sqrt{z^2 - \frac{1}{n} + 4n\hat{p}(1 - \hat{p}) + (4\hat{p} - 2)} + 1\right] }
{ 2(n + z^2) }$$
If these formulas give a value below $0$ or above $1$ (which is an impossible value for $p$) then round them to $0$ or $1$
This one doesn't have a nice way of adjusting for a small population size. If you have both a small population size and a small sample size I'd recommend prioritizing the small population size and using the second set of confidence interval formulas I described.
