Hi Nickos. Here is a possible answer.
Consider \(\displaystyle{M=\mathbb{R}\cup\,\left\{(0,1)\right\}}\) and the sets
\(\displaystyle{U_1=\left\{(t,0)\,,t\in\mathbb{R}\right\}\,\,,U_2=\left\{(t,0)\,,t\in\mathbb{R}\left\{0\right\}\right\}\cup\left\{(0,1)\right\}}\)
Also, define the maps
\(\displaystyle{\phi_1(t,0)=t\,,(t,0)\in U_1}\)
\(\displaystyle{\phi_{2}(t,0)=t\,,t\in\mathbb{R}\left\{0\right\}\,\,,\phi(0,1)=0}\).
We have that \(\displaystyle{M=U_1\bigcup U_2\,\,,U_1\bigcap U_2=\mathbb{R}\left\{0\right\}}\)
and \(\displaystyle{\phi_1\,,\phi_2}\) are homeomorphisms.
If \(\displaystyle{p=(0,0)\in M\,\,,q=(0,1)\in M}\), then for each \(\displaystyle{\epsilon>0}\) holds
\(\displaystyle{(\epsilon,\epsilon)\cap (\epsilon,0)\cup (0,\epsilon)\cup\left\{0\right\}\neq \varnothing}\)
and \(\displaystyle{M}\) is not a \(\displaystyle{\rm{Hausdorff}}\)  space.
