Let $f(x) = \sqrt{x}$. I'm trying to show that $f$ is continuous at $0$ by contradiction.
In order for a function to not be continuous at $0$: There exists an $\epsilon > 0$ such that for every $\delta_n = \frac{1}{n} > 0$, $|x_{n} - 0| \leq \delta_n \implies |\sqrt{x_n} - 0| > \epsilon$ where $x_n$ is an element in the domain of $f$.
Suppose $\epsilon = 1$. Then I see that $|\sqrt{x_n}| > 1 \implies x_n > 1$. But I see that $x_n \to 0$ since $\delta_n = \frac{1}{n} \to 0$.
This would be a contradiction because $x_n > 1$ which means that $\{x_n\}$ will never converge to 0.
Could I get some feedback on my proof?