Is my reasoning right? I have $f_n(x) = \sqrt{x^2 + \frac{1}{n^2}}$ for $x \in \mathbb{R}$, so I conclude that it's pointwise convergent $f_n \to |x|$, and moreover it's uniformly convergent to $|x|$, because $\left | \sqrt{x^2 + \frac{1}{n^2}}- |x| \right | = \frac{1}{n^2 \left ( \sqrt{x^2 + \frac{1}{n^2}} + |x| \right )} \leq \frac{1}{n^2} \to 0$
That's because $\sqrt{x^2 + \frac{1}{n^2}} \to |x|$ and it's decreasing for any $x \in \mathbb{R}$, therefore I can make denominator smaller by $\sqrt{x^2 + \frac{1}{n^2}} \leq |x|$, so I'd get in denominator $n^2 2|x|$ and if $|x| > \frac{1}{2}$ then $\frac{1}{n^22|x|} \leq \frac{1}{n^2}$, otherwise if $|x| \leq \frac{1}{2}$ then $\frac{1}{n^22|x|} \leq \frac{2|x|}{n^22|x|} = \frac{1}{n^2}$
A simpler way to proceed: If $a,b\ge 0,$ then $\sqrt {a+b}-\sqrt a \le \sqrt b.$ Proof: Move $\sqrt a$ to the other side and square. It follows that
$$\sqrt {x^2+1/n^2} -\sqrt {x^2} \le \sqrt {1/n^2} = 1/n.$$