Where is the Hausdorff condition used? Here https://en.m.wikipedia.org/wiki/Continuous_functions_on_a_compact_Hausdorff_space
It says that the space $C(\Omega)$ is a normed space if $\Omega$ is a compact hausdorff space. But why do we need the Hausdorff condition? Compactness gives that the sup norm for each function is finite. And the other normed properties seems to follow the same way they do if $\Omega=[0,1]$?
Also,  is the space $C(\Omega)$ complete? It seemed like it was when i mimicked the proof for the special case $[0,1]$. However wikipedia didn't mention completeness. And does this question depend on the hausdorff condition?
 A: It doesn't seem to be used for the construct of the space it self, but further down on the page there's results that seem to rely on it being a Hausdorff space, for example the consequence of Uhrysohn's lemma.
In fact a space where that consequence of Uhrysohn's lemma is valid (ie that points can be separated by a continuous function) has to be a Hausdorff space itself. Pick two points $x$ and $y$ then there's exsts a $f$ such that $f(x)\ne f(y)$ which means that you can form disjoint neighbourhoods of $f(x)$ and $f(y)$ whose inverse images are disjoint neighborhoods of $x$ and $y$.
The space is complete since if you have a cauchy sequence $f_n$ of such functions then it's obviously pointwise convergent to a function $f$. It's quite obvious that it will also be uniformly convergent to it and that $f$ is continuous (since $|f_j-f_k|<\epsilon$ for large enough $j$ and $k$ we have that $|f-f_j|<\epsilon$ as well and from the continuity of $f_j$ we can estimate $f$ so that it's also continuous).
A: As indicated in the other answers, the Hausdorff condition is not necessary, at least for the definition and most basic properties of $C(\Omega)$.  In fact, the compact condition is not necessary either as long as you change the definition to only include bounded continuous functions (often written $C_b(\Omega)$, with "b" for "bounded").  For any space $\Omega$, the sup norm makes $C_b(\Omega)$ a complete normed space, and in fact pointwise multiplication of functions further makes $C_b(\Omega)$ a Banach algebra.  A major reason that the article focuses on compact Hausdorff spaces is that for any space $\Omega$, there is an essentially unique compact Hausdorff space $\beta\Omega$ such that $C_b(\Omega)$ is isomorphic to $C(\beta\Omega)$ as a Banach algebra (and this isomorphism is implemented by composing with a canonical continuous map $\Omega\to \beta\Omega$).  This compact Hausdorff space $K$ is called the Stone-Čech compactification of $\Omega$.  So even if you are interested in $C_b(\Omega)$ for arbitrary spaces $\Omega$, you can often get away with only thinking about the case when $\Omega$ is compact and Hausdorff (at least, assuming you are capable of understanding the canonical map $\Omega\to \beta\Omega$ for the spaces $\Omega$ you care about, which is not always so easy...).
A: I agree that the proof of completeness of $C([0,1])$ applies without modifications to $C(\Omega)$ if $\Omega$ is any compact topological space. 
Actually, if $\Omega$ is any topological space, compact or not, the space of continuous and bounded functions defined on $\Omega$ and taking values in a Banach space is complete. The proof is still exactly the same.
I think that most of the theorems listed on the Wikipedia page use the Hausdorff condition (perhaps through Uryshon's lemma - as skyking points out). This is trivially true of the Stone-Weierstrass's theorem: if some points of $\Omega$ cannot be separated by open sets, any continuous function must be constant on them. So there is no subset of $C(\Omega)$ that separates points, and therefore Stone-Weierstrass's theorem is empty. 
However, I suspect that Ascoli-Arzela's theorem holds even without the Hausdorff property of $\Omega$. 
