Continuous functions on compact Hausdorff space. There is a well known theorem that says that if $X$ is a compact Hausdorff space, then the space $C(X)$ of the continuous functions on $X$ is a complete Banach space with the sup norm.
It's clear why the space $X$ should be compact: in this way the sup norm is well defined.
It's natural that the Hausdorff hypothesis is necessary to have a REASONABLE space, but why the Hausdorff hypothesis is necessary here?
Can you make a counterexample of the theorem failing without it?
 A: You don't need it for $C(X)$ to be a Banach space. But it guarantees that enough continuous functions exist to have nice theorems about the relation between $X$ and $C(X)$, eg. For $X$ any cofinite infinite space (so only compact $T_1$), the space $C(X)$ only consists of constant functions, so is isometrically isomorphic to $\mathbb{R}$, regardless of the cardinality of $X$. For compact Hausdorff spaces we have nicer theorems, and the algebraic structure of the ring $C(X)$ determines the topology of $X$ uniquely. So the theory is nicer when we add Hausdorffness. 
A: The Hausdorff property is not required.
If $X$ is compact, then for every $f\in C(X)$, $f[X]$ is a compact set in $\mathbb R$ or $\mathbb C$, and hence $\|f\|_\infty$ is well-defined. In fact
$$\|f\|_\infty=\max_{x\in X}\lvert f(x)\rvert.$$
In such case $C(X)$ is indeed a Banach space, as if $\|f_m-f_n\|_{\infty}\to 0$, then
$\lvert f_m(x)-f_n(x)\rvert\to 0$, for all $x\in X$, and thus $\{f_n(x)\}$ converges for every $x$, say to $f(x)$. It is not hard to show that $\|f_n-f\|_\infty\to 0$ - indeed for $\varepsilon>0$, there exists an $N\in\mathbb N$, such that $m,n\ge N$, implies that $\|f_m-f_n\|_\infty<\varepsilon$, thus $|f_m(x)-f_n(x)|<\varepsilon$, for all $x$ and taking the limit, as $m\to \infty$, we obtain  $|f_n(x)-f(x)|<\varepsilon$, for all $x$, and finally $\|f_n-f\|_\infty<\varepsilon$.
Finally, $f$ is continuous. Let $x_0\in X$ and $\varepsilon>0$ and $N\in\mathbb N$, such that $\|f_N-f\|_\infty<\varepsilon/3$. As $f_N$ is continuous at $x_0$, there exists a $U\subset X$, open, such that $x_0\in U$ and whenever $x\in U$, then $\lvert f_N(x)-f_N(x_0)\rvert<\varepsilon/3$. Hence, for every $x\in U$ we have that
$$
\lvert f(x)-f(x_0)\rvert\le \lvert f(x)-f_N(x)\rvert+\lvert f_N(x)-f_N(x_0)\rvert+\lvert f_N(x_0)-f(x_0)\rvert < \frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon.
$$
