proof of continuity of coordinate functions (Rudin) 

I find a hard time understanding the above example of continuity of coordinate functions. What I understood from the example is that a coordinate function takes a point in $R^k$ and gives a real number; how is that gonna be continuous on $R^k$. May someone explain please?
 A: $\phi_i$ just picks out the coordinate $x_i$. 
It's just the statement that if distance between two vectors goes to $0$, then the distance between each of their coordinates also goes to $0$.
If you want, you could use the definition of the standard metric on $\mathbb{R}^n$ and then plug in what everything is: 
$$ |\mathbf{x} - \mathbf{y}| = \sqrt{(x_1-y_1)^2 + ... + (x_n-y_n)^2 } $$ so 
$$|\phi_i(\mathbf{x}) - \phi_i(\mathbf{y})| = |x_i - y_i|= \sqrt{(x_i-y_i)^2}$$
As you can see, just as Rudin says, $$|\phi_i(\mathbf{x}) - \phi_i(\mathbf{y})| \leq |\mathbf{x} - \mathbf{y}|$$
And so, given $\epsilon$, put $\delta = \epsilon$, if $|\mathbf{x} - \mathbf{y}|< \delta$, then $|\phi_i(\mathbf{x}) - \phi_i(\mathbf{y})| < \epsilon$
A: The definition of continuity is the same as the one you are used to for single variate functions. If you want to keep distances in the range smaller than say $\epsilon>0$, then you can find a suitably small $\delta>0$ ball in the domain for which inputs in this ball map to outputs inside the $\epsilon>0$ ball. 
Try it with a two variable example: If 
$$
\phi_1(x,y)=x
$$
Then insuring $|x-x_0|+|y-y_0|\leq \delta$ (the 1 norm in the domain, norms on $\mathbb{R}^n$ are equivalent), then 
$$
|\phi_1(x,y)-\phi(x_0,y_0)|=|x-x_0|\leq \delta
$$
the norm in the range being used above.
So indeed, choosing $\delta=\epsilon$ meets any $\epsilon>0$ challenge.
