Solution: To show that $(X,\mathcal{T} )$ is path-connected, we need to give a path from $0$ to $1$. Consider the function \[ \gamma : [0,1] \to X, \ \gamma (t) = \begin{cases} 1, &\text{ if } t = 1 ;\\ 0, &\text{ if } \text{ otherwise}. \end{cases} \] It is clear that $\gamma (0) = 0$ and $\gamma (1) = 1$ and hence this is path from joining $0$ to $1$. It just remain to show that $\gamma $ is continuous. As we have three open sets, we will show that preimage of all these open sets are open in $[0,1]$ with the Euclidean subspace topology. Note that \begin{align*} \gamma ^{-1} (X) & = [0,1] , \text{ open by definition }. \\ \gamma ^{-1} (\emptyset) & = \emptyset , \text{ open by definition }. \\ \gamma ^{-1} (\{ 0 \} ) & = (0,1] = (0,2) \cap [0,1]. \end{align*} The last set is open in the subspace topology and hence $\gamma $ is a continuous path joining $0$ to $1$.