**E-D-I-T 1** @RasmusBentmann has pointed out to my error (in the $\,T_0\,$ case (I considered $\ f(X)\ $ instead of $X).\ $ Thus virtually nothing of my attempt is left. Perhaps Rasmus Bentmann can post his proof if it's ready. I'll have to think a bit more about the situation. Perhaps I'll remove my post unless I can provide a correct result, either one pretty soon.

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**E-D-I-T 2** My "proof" was false. **Eric Vofsey** has provided a correct proof though in a comment below--I hope that Eric will his argument into an *Answer* (then I'll remove my answer immediately upon seen it, and with a relief).

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**THEOREM** $\ \left|Cont(X\ X)\right|\ \ge\ |X|+2\,\ $ for every finite topological space $\ X\ $ such that $\ |X|\ge 3$.

**PROOF** We need to show that there is a continuous function $\ f:X\rightarrow X\ $ such that it is not the identity, nor constant.

First, let $\ X\ $ be not a $T_0$-space, i.e. there are two different points $\ a\ b\in X\ $ such $\ \left|G\cap\{a\ b\}\right|\ne 1\ $ for arbitrary open $\ G\subseteq X.\ $ Define $\ f:X\rightarrow X\ $ by conditions:

- $\ f\,|\,X\!\setminus\!\{a\}\ $ is the identity on $\ X\!\setminus\!\{a\}$
- $\ f(a)\, :=\, b$

Then obviously $\ f^{-1}(G)\,=\,G\ $ for every open $\ G\subseteq X.\ $ Thus $\ f\ $ is continuous, it's not an identity on $\ X,\ $ and it's not a constant map (since $\ |X|\ge 3$. Thus the theorem holds in this case.

Now let $\ X\ $ be a $T_0$-space. Then see the **Eric**'s argument in the comment below (until and if Eric expands his comment into an *Answer*).