Ask yourself if the spaces $C^k(0,1)$ are of interest.
EDIT: I also suggest thinking more about and therefore formalising the notion of a 'good' metric on a space. You might find that a good metric is one that induces a topology on the space that has desirable properties for doing analysis, like local convexity, being complete, the Heine-Borel property, an interesting dual space, etc.
Since $(0,1)$ is the union of countably many compact sets $K_n$ such that $K_n\subset\text{int}K_{n+1}$, then we define $C(0,1)$ as the space of all continuous real-valued functions on $(0,1)$ with topology induced by the family of separating semi-norms:
$$
p_n(f):=\sup_{x\in K_n}|f(x)|.
$$
Then the sets
$$
V_n:=\{f\in C(0,1)\;|\;p_n(f)<1/n\},\qquad n=1,2,...
$$
form a local convex base for $C(0,1)$. This topology is in fact induced by the metric
$$
d(f,g):=\max_n\frac{1}{2^n}\frac{p_n(f-g)}{1+p_n(f-g)}
$$
This metric is complete, so $C(0,1)$ becomes a Frechet space.
The space $C^\infty(0,1)$ is defined to be the space of all real-valued functions $f$ on $(0,1)$ with the property that $D^kf\in C(0,1)$ for all $k\in\mathbb{N}_0$. Choosing the same compact sets $K_n$ as above, the following family of semi-norms undices a metrizable, locally-convex topology on $C^\infty(0,1)$:
$$
p_n(f):=\max_{x\in K_n,\;k\leq n}|D^kf(x)|.
$$
The metric on this space is of the same form as above. It can be shown that this space is also a Frechet space and additionally has the Heine-Borel property, so it follows that it is not normable (the metric is not induced by a norm). On the space $C^\infty(0,1)$ is constructed the space of distributions with compact support, which are linear functionals that are continuous with respect to the above defined topology.
To generalise the above, you should replace $(0,1)$ with an arbitrary open subset of $\Omega\subset\mathbb{R}^m$, and allow $k$ to be a multi-index.
Other continuous function spaces that you might be interested in are:
(i) the space $\mathcal{D}_K$ which is the space of all $f\in C^\infty(\mathbb{R}^m)$ such that $\text{supp}f\subset K$ where $K\subset\Omega$ is compact. It can be shown that $\mathcal{D}_K$ is a closed subspace of $C^\infty(\Omega)$.
(i) the Schwartz space $\mathcal{S}(\mathbb{R}^m)$, which is important for Fourier analysis. On this space is consructed the space of tempered distributions, which is composed of linear functionals that are continuous with respect to the appropriate topology defined on the Schwartz space.
(ii) the space of test functions $\mathcal{D}(\Omega)=C^\infty_0(\Omega)$, which is important for Sobolev space and PDE theory. The space of distributions is constructed on this space, and consists of all linear functionals that are continuous with respect to the appropriate topology on $\mathcal{D}(\Omega)$. This topology is a challenge to define and understand, but is easily characterised in terms of convergence.
By far the best reference for all of this stuff is Rudin's book on functional analysis. Also see Yosida's book on functional analysis, DiBenedetto's book on real analysis and Knapp's book on advanced analysis.
Best Answer
It has become little bit lengthy because I tried to avoid the use of sequences of functions.
We apply the general definition of a limit point:
"$f$ is a limit point of the set $F$ iff for all $\epsilon>0$ there exists a $g\in F$ with $f\neq g$ such that $\Vert f-g\Vert <\epsilon$".
Now we take an arbitrary limit point $f$ of $F$ and show that $f<0$ is impossible.
Let's assume $\epsilon>0$ and $f(z)=-\epsilon$ for an arbitrary $z\in[0,1]$. As stated above, we know that there exists a $g\in F$ with $g\neq f$ and $\Vert f-g\Vert = \sup\limits_{x\in[0,1]}|f(x)-g(x)| <\epsilon$. Note that $g(x)\geq 0$ for all $x\in[0,1]$. This leads to a contradiction because then we would have $\epsilon<|-\epsilon-g(z)|=|f(z)-g(z)|\leq\sup\limits_{x\in[0,1]}|f(x)-g(x)| <\epsilon$. So it must hold that $f(x)\geq 0$ for all $x\in[0,1]$ as $z$ was arbitrarily chosen.
Now we check that $f$ is continuous:
We know that every $g\in F$ is continuous on a closed intervall and hence uniformous continuous. So we know that for a given $\epsilon>0$ there exists a $\delta>0$ such that for all $x,y\in[0,1]$ with $|x-y|<\delta$ it holds $\vert g(x)-g(y)\vert<\epsilon$.
Let's choose an arbitrary $\epsilon>0$ and define $\epsilon':=\frac{\epsilon}{3}>0$. So there exists a $g\in F$ with $\Vert g-f\Vert <\epsilon'$. Moreover we know that there exists a $\delta>0$ such that for every $x,y\in[0,1]$ with $|x-y|<\delta$ it holds $\vert g(x)-g(y)\vert<\epsilon'$ (uniformous continuity of $g$). Let be $a\in[0,1]$ arbitrary but fixed and $x\in[0,1]$ with $|x-a|<\delta$. Then we have $$\vert f(x)-f(a)\vert \leq \vert f(x)-g(x)\vert +\vert g(x)-g(a)\vert+ \vert f(a)-g(a)\vert \leq \\ \Vert f-g\Vert +\vert g(x)-g(a)\vert+ \Vert f-g\Vert<\epsilon'+\epsilon'+\epsilon' = \epsilon.$$
So $f$ is continuous on $[0,1]$ because for an abritrary $\epsilon>0$ we have found a $\delta>0$ such that for all $x\in[0,1]$ with $|x-a|<\delta$ it holds that $|f(x)-f(a)|<\epsilon$.
So we can conclude that $f\in F$. As $f$ was arbitrarily chosen $F$ is closed.