I have divide the interval $[0,1]$ into $k$ equal sub-intervals, which I call classes, and generated $n$ observations from a uniform distribution. The number $X_{1}$ of the $n$ observations that fall in the first interval is a random variable, and the number of $X_{2}$ which ends up in the second interval and so on. I take it that number of observations are binomial distributed in the sub-intervals, so $X_{j} \sim Bin(n,\frac{1}{k})$ for $j=1,\dots,k$.
Consider now the proportion $Y_{1} = \frac{X_{1}}{n}$ of observations that fall into the first class, which also is a random variable.
Is it sufficient to simply divide the expected value, variance and standard deviation for $X_{1}$, which is binomial distributed, by $n$ to obtain the corresponding values for $Y_{1}$?
That is to say, is $\mathbb{E}(Y_{1})=\frac{1}{k}$, $Var(Y_{1})=\frac{k-1}{k^{2}}$ and $\sigma=\sqrt{\frac{k-1}{k^{2}}}$?
$X_{1}$ has binomial distribution with parameters $n$ and $p=\frac{1}{k}$. Can you take it from here?
addendum:
For constant $c$: $\mathbb{E}\left(cX\right)=c\mathbb{E}X$ hence: $\mathbb{E}\left(cX\right)^{2}=\mathbb{E}c^{2}X^{2}=c^{2}\mathbb{E}X^{2}$.
So:$Var\left(cX\right)=\mathbb{E}\left(cX\right)^{2}-\left(\mathbb{E}\left(cX\right)\right)^{2}=c^{2}\mathbb{E}X^{2}-\left(c\mathbb{E}\left(X\right)\right)^{2}=c^{2}\left(\mathbb{E}X^{2}-\left(\mathbb{E}X\right)^{2}\right)=c^{2}Var\left(X\right)$.
Then: $\sigma\left(cX\right)=\sqrt{Var\left(cX\right)}=\sqrt{c^{2}Var\left(X\right)}=\left|c\right|\sqrt{Var\left(X\right)}=\left|c\right|\sigma\left(X\right)$.