Trying to find sample standard deviation of a proportion The data set is 32% of n:2000
CI: 95
I did
$x=\pm 1.96\sqrt{\frac{0.32*0.68}{2000}}$
and
$CI=0.32\pm 1.96 * \frac{0.466}{\sqrt{2000}}$
to find the confidence interval
My numbers came out pm 0.34, 0.29

The interval you are using is called the Wald interval, and it is not a particularly good method for estimating a proportion in binary data.  A better approach is to use the Wilson score interval, specifically designed for estimating a proportion in binary data.  (The two methods give similar answer in this case because you have a large sample size, but in general, the Wilson score interval is far superior.)
It is simple to implement the confidence interval for a proportion using the CONF.prop function in the stat.extend package (in the statistical computing language R).  Here is how to do it for the data you have:
library(stat.extend)
CONF.prop(alpha = 0.05, n = 2000, sample.prop = 0.32)

        Confidence Interval (CI) 
 
95.00% CI for proportion parameter for infinite population 
Interval uses 2000 binary data points with sample proportion = 0.3200 

[0.299917915975882, 0.340772221045369]

