I have the following question:
In the following link, $f_1,f_2,\ldots$ continuous on $[0,1]$ s.t $f_1 \geq f_2 \geq \ldots$ and $\lim_{n\to\infty}f_n(x)=0$ ., a proof was provided which did not seem to require the Hausdorff property.
I.e. The following proof is a simple adaptation of the one given by Michael Hardy.
define $C_{n,\varepsilon} = \{x\in X : f_n(x) \ge \varepsilon\}=f_n^{-1}([\epsilon,\infty))$. Clearly $C_{1,\varepsilon} \supseteq C_{2,\varepsilon}\supseteq\cdots$ by the decreasing condition. Each $C_{n,\varepsilon}$ is closed since each $f_n$ is continuous. Also the intersection of all the sets $C_{n,\varepsilon}$ has to be 0 since $f_n$ goes to $0$ for each $x$. Hence by the finite intersection property, the intersection of some finite subcollection is empty. Hence there exists $n$ s.t $C_{n,\varepsilon}$=$\emptyset$. This implies uniform convergence.
Does this mean that the Hausdorff property is not required.