The closure of a singleton is path-connected Let $x$ be in an element of topological space $X$. Prove that the closure of $\{x\}$ is path connected.
I am having issues creating a path, and I have a bit of difficulty understanding "the closure of $\{x\}$". 

Take any point $y\in \overline{\{x\}}$ and consider the curve $\gamma\colon[0,1] \to X$ defined by
$$
\gamma(t) = \begin{cases}y &\text{when $t=0$}\\x &\text{when $t>0$}\end{cases}
$$
it is enough to prove that $\gamma$ is continuous.
For $t>0$ the function is constant hence continuous. For $t=0$ take any neighbourhood $U$ of $y$. Since $y\in \overline{\{x\}}$ then $U\cap\{x\}\neq \emptyset$ hence $x\in U$. So $\gamma([0,1]) = \{x,y\} \subset $U$.
