Elementary question on pointwise convergence and norm continuity 
Let $\;a_n \in \mathbb R\;$ be a bounded sequence, then by 
  Bolzano-Weierstrass Theorem it follows there exists a subsequence 
  $\;a_{n_k}\subset a_n\;$ such that $\;a_{n_k} \to a \in \mathbb R\;$. 
  If $\;f:\mathbb R \to \mathbb R^m\;$ a continuous function, then
  $\;f(x-a_{n_k}) \to f(x-a)\;$ pointwise.
In addition, $\;\forall g \in C(\mathbb R;\mathbb R^m)\;$ it holds:
  $\;{\vert g(x)-f(x-a_{n_k}) \vert}^2 \to {\vert g(x)-f(x-a) \vert}^2\;$ also
  pointwise.

NOTE: $\; \vert \cdot \vert \;$ stands above for the Euclidean norm
As I was studying, the above part of my notes confused me a little bit. 
QUESTIONS:


*

*Why is $\;f(x-a_{n_k}) \to f(x-a)\;$ pointwise? I know that uniform convergence is much stronger than pointwise but why is that the case here? 

*Why does the second convergence hold? Is it because $\;\vert \cdot \vert\;$ is also continuous?


Any help would be valuable. Thanks in advance!

Consider $y_{n_k}=x-a_{n_k}$ and $y =x-a $ for any fixed $x\in \mathbb{R}$, then $y_{n_k} \rightarrow y $ in $\mathbb{R}$. Now as both $f$ and $g$ is continuous in $\mathbb{R},$ you can have both the convergence. Also you have to use the fact $||$ is continuous and composition of two continuous maps is continuous  to get the last result. 
