Show that $f$ is continuous at $a$ if every $f_i$ is continuous at $a$ 
Show that $f=(f_1,f_2, \dots, f_m): \mathbb{R^n} \to \mathbb{R^m}$ is continuous at $a$ iff each of the component functions of $f$ are continuous at $a$.

If I suppose that every $f_i$ is continuous then I would get that for all $ \varepsilon >0$ there is $\delta >0$ s.t. $|f_i(x)-f_i(a)| < \frac{\varepsilon}{\sqrt{n}}$ when $|x-a| < \delta.$
$$|f(x)-f(a)|^2 = \sum_{i=1}^n|f_i(x)-f_i(a)| < n(\frac{\varepsilon}{\sqrt{n}})^2 = \varepsilon^2.$$
However I feel like I'm assuming the thing here that I want to proof. Any hints on how should I correct this?
 A: Your $\delta$ depends on $i$. Call it $\delta_i$ and take $\delta =\min \{\delta_1,\delta_2,..., \delta_n\}$. (There is also a typo: you missed  square).
For the converse use the fact that $f_i$ is the composition of $f$ with the continuous map $(x_1,x_2,...,x_m) \to x_i$.
A: Fixing your argument:
Let $\varepsilon>0$.
If every $f_i$ is continuous, then for every $i=1,\ldots,n$, there is $\delta_i >0,$ s.t. $|f_i(x)-f_i(a)| < \frac{\varepsilon}{\sqrt{n}}$, when $|x-a| < \delta_i.$
Setting
$$
\delta=\min\{\delta_1,\ldots,\delta_n\},
$$
we have
$$
|f(x)-f(a)|^2 = \sum_{i=1}^n|f_i(x)-f_i(a)| < n(\frac{\varepsilon}{\sqrt{n}})^2 = \varepsilon^2.
$$
whenever
$$
|x-a|<\delta.
$$
A: Because all norms are equivalent on $\Bbb R^n$ we are free of choice which one we use. So let's take $$||x||_\max = \max_{i} |x_i|$$ and we need to show: $$||f(x) - f(a)||_\max < \varepsilon \iff |f_i(x) - f_i(a)| < \varepsilon \quad \forall i$$ whenever $|x-a| < \delta$. But this is true by definition of $||x||_\max$ and we are done.
