The image
$d\colon C([a,b])\times C([a,b])\to \mathbb{R}_{\ge 0}\\ (f,g)\mapsto \{\sup |f(t)-g(t)|,t\in [a,b]\}$
defines a metric on the set of all continuous functions. I want to show that
$\Phi_x\colon C([a,b])\to \mathbb{R}\\ f\mapsto f(x)\\ x\in[a,b]$
is continous. I used the absolute value metric on the field of real numbers.
For an arbitrary $\epsilon >0$ I can find a $\delta >0$ that
$|f(x)-g(x)|\le \sup |f(t)-g(t)| <\epsilon$ holds if I choose $\epsilon = \delta$, thus the function is continuous.
Is this correct?
In the sup metric it's quite obvious that for any $x \in [a,b]$: $|\Phi_x(f)| \le \sup\{|f(t): t \in [a,b]\}= ||f||_\infty$, as the sup of a set is an upperbound for it.
As we have a linear map $$|\Phi_x(f) - \Phi(g) | = |\Phi_x(f-g)| \le || f-g||_\infty = d(f,g)$$
So for any $\varepsilon > 0$ and any $g$, we show continuity of $\Phi_x$ at $g$ by picking $\delta = \varepsilon$, then by the above inequality $d(f,g) < \delta$ implies $|\Phi(f) - \Phi(g)| < \delta = \varepsilon$ as well.