Proving a simple functional is continuous I recently took real analysis at my school, and enjoyed it thoroughly. I decided to use some of my Summer to study variational calculus, and wanted some verification of my work (or correct work if I'm wrong) for this example.

Let the functional $\phi$ be defined by:
  $$\phi[h] = h(x_0) \space\text { for each function  } \space h(x) \in \mathbb{F}(a,b) $$
  Where $\mathbb {F}(a,b)$ is the set of continuous functions on the interval $(a,b) $ and $x_0 \in (a,b) $
Prove that $\phi [h] $ is a continuous linear functional on this function space.

The linearity part was trivial, I just want verification in prove continuity.
$\phi $ is continuous at $h(x)$ provided that $\forall \epsilon >0, \exists \delta > 0 \text { s.t. }$
$$||h(x)-g(x)||<\delta \implies |\phi[h]-\phi [g]|<\epsilon$$
Where the norm on $\mathbb{F}(a,b) $ is defined as:
$$||h(x)|| = \max_{a \le x \le b} |h(x)|$$.
My thought was that if we let $\delta = \frac{||h(x)-g(x)||}{|h(x_0) - g (x_0)|}\epsilon $,
Then we are done. 
Is this an acceptable approach? I thought it would work, since $\delta $ is only a function of $\epsilon $ for any given $g $, but if we were to adapt it to a general case, this argument could prove (incorrectly) that anything is continuous, so I'm very wary of this style. 
Could someone explain:
1) where my argument fails if it is incorrect, and provide a correct one
2) where this style argument would fail for a general functional if it is correct
 A: So, here is an $\epsilon$ - $\delta$ proof.
First, as Simonsays suggests, we should work with continuous functions on a closed interval, in order for the norm to be well defined. Alternatively, we should consider only bounded functions. This way or another, let $h\in \mathbb{F}(a,b)$, and let $\epsilon>0$. Take $\delta=\epsilon$. For every $g\in\mathbb{F}(a,b)$ satisfying $\|g-h\|<\delta$, we have$$|\phi(g)-\phi(h)|=|g(x_0)-h(x_0)|\leq\max_{x\in[a,b]}|g(x)-h(x)|=\|g-h\|<\delta=\epsilon.$$ The major difference between this argument and the one in the question, is that here, $\delta$ does not depend on $g$.
A: I assume now that you work on compact $[a,b]$ and you can  use the following: a linear functional $\phi$ is continuous if and only if its operator norm is bounded, which means
$ ||\phi|| := sup_{||h||=1,h \in \mathbb{F} }|\phi(h)| < \infty$. So if a functional is continuous, it is already Lipschitz continuous, with Lipschitz constant $||\phi||$. In that specific application $||\phi||$ is clearly bounded, since $|\phi(h)|=|h(x_0)| \leq max_{x \in [a,b]}|h(x)| = 1$ for $x_0 \in [a,b] $ and $||h||=1$.
