If in a normed space $a_n$ converges to $a$, prove that $||a_n||$ converges to $||a||$ If in a normed space an converges to $a$, prove that $||a_n||$ converges to  $||a||$

I have seen this proof: for all $\varepsilon&gt;0$ there exists and $N$ s.t. $n&gt;N$ implies
that the $||a_n-a||&lt;\varepsilon$

by the reverse triangle inequality for the same $N \varepsilon &gt; ||an-a|| &gt; |\ ||an||-||a||\ |$ 

and therefore it converges.  

But this is the absolute value of the difference of norms.  Don't you need the norm of the difference of norms to be less than $\varepsilon$?

As in

$||\ ||an|| - ||a||\ || &lt; \varepsilon$

not

$|\ ||an|| - ||a||\ | &lt; \varepsilon$

Or am I confused

Thanks!
The expressions $||a_n||$ and $||a||$ are real numbers, rather than elements of your normed vector space (unless the normed vector space is $\mathbb{R}$ with the usual absolute value, in which case $||\cdot||=|\cdot|$), so you need to look at the absolute value of the difference, rather than the norm of the difference.
Let your normed space be $X$. The map $||\cdot|| : X \to \mathbb R$ is continuous. Since $a_n \to a$, $||a_n|| \to ||a||$, as the norm is continuous, and continuous functions preserve sequential convergence, limits being sent to limits.