[From 'Counterexamples in Topology']1
Let $\tau$ be the supremum topology of the usual topology $\tau_1$ and the cocountable topology $\tau_2$; show that $(\mathbb{R}; \tau)$ is not locally connected.
Considering $A=]-\infty;1[-\{\frac{1}{n} \mid n \in \mathbb{N}\}$; we have $\mathscr{C}_0=]-\infty;0] \notin \tau$ as if $\mathscr{C}_0=]-\infty;0]$ were in $\tau$:
$$]-\infty;0]=U-C; \ \textit{U open; C countable} $$ then $]-\epsilon;\epsilon[ \subset U$ for some $\epsilon >0$ but $]0; \epsilon[-C\neq\emptyset$. So $]-\infty;0]$ is not open and there doesn't exist any connected neighborhood of $0$ that is a subset of the open neighborhood $A$, which leads to the conclusion that there is no local basis for $0$.
Which argument should I use in order to show that $]-\infty;0]$ is connected ?
