Prove $f$ is continuous at $x = 0$ in the topological way

I want to prove that $f: [0, +\infty) \to \mathbb{R}$ defined by $f(x)= \sqrt{x}$ is continuous at $x = 0$ in the topological way, that is (from Wikipedia): $F: X \to Y$ is continuous at $x = x_0 \in X$ if for every neighbourhood $V$ of $F(x)$ in $Y$, there exists a neighbourhood $U$ of $x$ in $X$ such that $F(U) \subseteq V$.

Attempts

So $x = 0 \in X$. I chose a neighbourhood of $0$ in $X$, like $U = [0, 1/2)$. Then $f(U) = [0, \sqrt{1/2})$.

Does it suffice then to say I can choose, for every $\delta > 0$, $V$ such that $V = [0, \sqrt{1/2} + \delta)$?

In this way I have $f(U) \subseteq V$.

I don't know, I think it's not correct and not that simple...