TITLE: Is there $(X,\mathcal{T})$ contractible, $\mathcal{T}'\subseteq\mathcal{T}$ such that $(X,\mathcal{T}')$ isn't contractible?
QUESTION [5 upvotes]: Let $(X,\mathcal{T})$ be a contractible space and let $\mathcal{T}'\subseteq\mathcal{T}$ be a coarser topology on $X$ than $\mathcal{T}$. Is it always true that $(X,\mathcal{T}')$ is also contractible?
I highly doubt that this is the case, because intuitively, if it were true, then it should work with the same homotopies for $\mathcal{T}'$ as for $\mathcal{T}$ because of the generality of the statement (I mean, given the generality, we cannot adapt the homotopy equivalences for $(X,\mathcal{T})$ in a way that makes sense), but there is absolutely no reason for $F:X\times I\to X$ to still be continuous if we equip $X$ with $\mathcal{T}'$ instead of $\mathcal{T}$.
However, I fail to come up with a counterexample; the eight topologies I know over $\mathbb{R}$ don't provide one. Can you give an elementary example where this fails?

REPLY [4 votes]: Stefan beat me to the (same) answer, but I think it's worth having a different explanation about the contractibility, using the homotopy theory of finite spaces.  

Playing around with finite spaces, I think I've found an example.  Unfortunately, I don't think it occurs "in practice".  
Let $X = \{a, b, c, d\}$ be a four point set.  Let $\mathcal{T}$ be the topology generated by the basis $\{a\}$, $\{b\}$, $\{a, b, c\}$, and $\{b, d\}$, and let $\mathcal{T}'$ be the topology generated by the basis $\{a\}$, $\{b\}$, $\{a, b, c\}$, and $\{a, b, d\}$.  It is easy to verify that $\mathcal{T}' \subsetneq \mathcal{T}$.  
I claim that $(X, \mathcal{T})$ is contractible.  There is a classification of the homotopy type of finite spaces by their cores.  See chapter 2 of Peter May's finite spaces book for an explanation of this and the terminology I am going to employ.  Anyway, for $(X, \mathcal{T})$, the point $d$ is a downbeat point, witnessed by $b$.  So there is a deformation retract of $(X, \mathcal{T})$ onto $(\{a,b,c\}, \mathcal{T}|_{\{a,b,c\}})$.  But this latter space is the (non-Hausdorff) cone on $S^0 = \{a,b\}$.  So it is contractible.  Another way to see this is that $c$ is a maximum element of the associated poset.  
On the other hand, $(X, \mathcal{T}')$ is the (non-Hausdorff) suspension on $S^0 = \{a,b\}$, which is weakly homotopy equivalent to $S^1$.  So it is not contractible.  Alternatively, note that $(X, \mathcal{T}')$ is minimal finite, so is its own core.  Since its core is not a point, $(X, \mathcal{T}')$ is not contractible.