Hölder continuity for operators

Your proposed inequality doesn't even work on diagonal matrices. Let $x\in\mathbb M_n$ be diagonal with entries $x_i\geq 0$. Then $x^{1/2}$ is diagonal with entries $\sqrt{x_i}$. Thus $$ \|x^{1/2}\|_{HS}^2 = \sum_i x_i, \qquad \|x\|_{HS} = \Big( \sum_i x_i^2 \Big)^{1/2}. $$ Taking e.g. $x_i=1/\sqrt n$ we obtain $n/\sqrt n \leq C^2$ so $C\geq n^{1/4}$.

The Powers-Stormer inequality (see section 4 of https://projecteuclid.org/euclid.cmp/1103842028) says that $$ \|x^{1/2} - y^{1/2}\|_{HS}^2 \leq \|x - y\|_1 $$ for all $x,y$ positive and Hilbert Schmidt. Here $\|\cdot\|_1$ is the trace-class norm (which takes the value $\infty$ if $x-y$ is not trace-class).

In finite-dimensions, we have that $\|x\|_1 \leq n^{1/2} \|x\|_{HS}$ (reduce to the diagonal case and apply Cauchy-Schwarz) and so we see that $C=n^{1/4}$ works in general.

For the Hilbert-Schmidt norm, the inequality $\|X^{1/2}-Y^{1/2}\|_2 \le C\|X-Y\|_2^{1/2}$ is false in general. Consider for that the case of $n\times n$ positive definite matrices. Put e.g., $Y=0$, then asking for a $C>0$ amounts to requiring \begin{equation*} \text{tr}\,X \le C'\sqrt{\text{tr}\,X^2} \end{equation*} which in general cannot hold for a constant independent of the dimension of $X$.

However, for all positive operators $X$ and $Y$, and for any unitarily invariant norm $|\!|\!| \cdot |\!|\!|$ \begin{equation*} |\!|\!|X^p - Y^p|\!|\!| \le |\!|\!| |X-Y|^p |\!|\!|,\quad 0 \le p \le 1. \end{equation*} Thus, in particular for the operator norm, this inequality implies $p$-Hölder continuity (with $C=1$).