Show that $f(x) = |x|$ is continuous on $\mathbb R$ I want to use the sequential criterion to prove that $f(x) = |x|$ is continuous on $\mathbb R$.
For reference, here is the sequential criterion according to Introduction to Real Analysis by Bartle:

$f:A \rightarrow \mathbb R$  is continuous at the point $c\in A$ if and only if for every sequence $(x_n)$ that converges to $c$, the sequence $(f(x_n))$ converges to $f(c)$.

So my attempt was this:
Let $x  \in \mathbb R$ and define the sequence $(x_n) = x + \frac{1}{n}$ for $n \in \mathbb N$ so $(x_n)$ converges to $x$. Then $(f(x_n)) = |x_n| = |x + \frac{1}{n}|$ so $(f(x_n))$ converges to $|x|$. Since this is true for all $x \in \mathbb R$, therefore $f(x) = |x|$ is continuous on $\mathbb R$.
My main source of doubt comes from that it seems you can show a lot of functions are continuous simply by slapping on a "$+\frac{1}{n}$". 
Is this proof valid? Or is there a constraint that I am not meeting?

You can make this easier if you just split the -function into cases: $|x|=x$ if
$ \geq0$ and $|x|=-x $ otherwise. Then for the first case, what you need to show is, essentially, tautological : $ (x_n \rightarrow x ) \rightarrow (x_n \rightarrow x)$ , since $f(x)=x $ in this case. Essentially the same for $f(x)=-x $.  
