Can I have help:
let $\displaystyle X$ be Uncountable set with cofinite Topology. Show that every continuous function $\displaystyle X \to R^1 $ is constant.
Here's the simplest possible way I can think to explain it. Suppose that $\displaystyle f(x)\ne f(y)$ for some $\displaystyle x,y\in X$ we can then separated the two by open sets (since $\displaystyle \mathbb{R}$ is Hausdorff) say $\displaystyle f(x)\in U,f(y)\in V$ and $\displaystyle U\cap V=\varnothing$. We have then that $\displaystyle f^{-1}(U)\cap f^{-1}(V)=\varnothing$ and $\displaystyle f^{-1}(U)\cap f^{-1}(V)$ are open and non-empty. That said, note that if $\displaystyle f^{-1}(U)\cap f^{-1}(V)=\varnothing\implies \left(X-f^{-1}(U)\right)\cup (X-f^{-1}(V))=X$, but we know that both $\displaystyle X-f^{-1}(U),X-f^{-1}(V)$ are finite since the only open sets in $\displaystyle X$ are $\displaystyle X,\varnothing$ and the complement of finite sets, and so if $\displaystyle X-f^{-1}(U)$ is infinite then $\displaystyle X-f^{-1}(U)=X$ which implies that $\displaystyle f^{-1}(U)=\varnothing$ contradictory to assumption. Thus, $\displaystyle (X-f^{-1}(U))\cup (X-f^{-1}(V))$ is the union of two finite sets, and so finite, but this is ludicrous since their union is $\displaystyle X$.
Note more generally this proves that any map $\displaystyle X\to Y$ where $\displaystyle X$ is infinite and $\displaystyle Y$ Hausdorff is constant.