The standard deviation must be positive - should you take the negative square root of the variance, it will be negative and thus would not make sense.
Now you certainly know (at least from the post you mentioned) that an unbiased estimate of the population variance, computed from a set of observations from this population $x_1, \dots, x_n$, is $S^2=\frac{1}{n-1}\sum_{i=1}^n (x_i-\bar x)^2$. Unbiasedness means $E[S^2] = \sigma^2$. Both are positive numbers, so you can take the (positive) square root of either side and the equality still holds,
\begin{align*} \sqrt{E[S^2]} &= \sqrt{\sigma^2}\\ &\Leftrightarrow \\ \sqrt{E[S^2]} &= \sigma \end{align*}
Where $\sigma$ is the standard error. Now, by Jensen's inequality, since the square root is a concave function, you have
$$ \sigma = \sqrt{E[S^2]} \geq E[\sqrt{S^2}] $$
so you actually systematically under-estimate the standard error.
[Edit] I just saw that the argument was given in another answer from the post you linked. The original poster defined the bias of an estimator $s$ of $\sigma$ as $\sigma - E[s]$ rather than $E[s] - \sigma$.