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?

This is not appropriate. Notice that the sequential criterion says that for EVERY sequence that converges to c. 
You have only looked at a single sequence, the sequence $x+1/n$. 
Instead, you would have to start your proof "Let $(x_n)$ be a sequence that converges to $(c)$. Then, from this, use the definition of sequence convergence to derive that $f(x_n)$ converges to $f(c)$. 
Edit: By your approach, the function $f(x) = 0,$ if x is rational and 1, if x is irrational, would be continuous at all rational points! That's a problem, isn't it?
