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Let $C([a,b])$ be the space of all continuous real functions on the compact interval $[a,b]$. This space shall be endowed with the topology of pointwise convergence, which basis is given by

$\mathcal{B}=\{x_1,...,x_n;t_1,...,t_n,\epsilon_1,...,\epsilon_n| f(x_i)\in B_{\epsilon_i}(t_i),i=1,...,n\}$ where $B_\epsilon(t)$ is an epsilon ball around $t$ with radius $\epsilon$.

I was trying to obtain the connected components of $C([a,b])$.

Firstly, I figured that the space must be connected, because I can write it as a union of disjoint open sets, something like this $C([a,b])=\bigcup_{i\in \mathbb{N}}\mathcal{B}_i$. At least i hope i am correct with this.

If I wanted to try to write down the connected components of e.g. $g(x)\in C([a,b])$ I can be sure, that it could be found in at least one $\mathcal{B}$, where the values on the finite $x_i$ coincide for $f$ and $g$. If I were to choose another set of finite values at $x_i$ where $f$ and $g$ coincide then two sets $\mathcal{B}$ would contain $g$ and thus can not be disjoint. So the only connected component is $g$ itself. Is this correct?

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    Every space is a union of open sets. Doesn't make it connected. A disconnected space is one that can be written as the union of disjoint open sets, after all. – 2017-02-25
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    Fixed. I am not sure if the set C(a,b) is connected at all. – 2017-02-25
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    I think your description of $\mathcal{B}$ is a little wonky. It should be a set of sets of functions. I interpret it as the statement that for every $n$ and every triple of $n$-tuples $(x_i)_1^n, (t_i)_1^n, (\epsilon_i)_1^n$, the set of functions satisfying $f(x_i)\in B_{\epsilon_i}(t_i)$ for each $i=1,\dots,n$ is an open set of the basis. But in this case, I don't see why to think that $C([a,b])$ can be written as a disjoint countable union of these. – 2017-02-25
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    Isn't $C([a,b])$ path-connected, and hence connected? – 2017-02-25
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    You interpreted this correctly, it is meant like this. I thought about it, and I think, as you suggest, $C([a,b])$ can indeed not be written as a disjoint open union. To be countable is not necessary i think by the way. – 2017-02-25

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Hint: Let $f: [a,b] \to \mathbb R$ be any continuous functions.Show that $H(x,t)=tf(x)$ gives a path between zero function and $f$.

Hence $C[a,b]$ is path connected...

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    I now know that the set is connected, but what are the connected components of C? – 2017-02-25
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    What happens when you look up the definition of connected components and apply that definition to a connected set? Do you reach any conclusions? – 2017-02-25