Is $\sqrt{x}$ continuous at $0$?

Since the function is not defined on the left of 0 then the notion of continuity in 0 consists only of the continuity at the right of 0.

Everyone will agree that the function $g: \Bbb R \to \Bbb R$

$$ g(x) = \left\{\begin{array}{lr} 0, & \text{for } x \lt 0 \\ \sqrt x , & \text{for } x \ge 0 \end{array}\right\} $$

is continuous.

Restrict this function to the domain $[0,+\infty)$, defining

$\tag 1 \displaystyle{f = g_{\;|\,[0,+\infty)}}$

We leave it to OP to come up with sensible terminology/definitions concerning continuity for the function $f$. But if you want the restriction of a continuous function to also be continuous (seems reasonable), you won't have much 'leg room'.