How to show that this space is not path connected? I apologize if this question has been already asked, but I can't find this space anywhere.
$$X = \left\{\left(0.5, 1\right] \times \{0\} \; \cup \; \bigcup_{n \in \mathbb{N}} \left\{ \left(x, \frac{x}{n} \right): x \in [0,1] \right\}\right\} $$
How can I prove that this space is not path connected? Also, how would I approach local path connectedness?

Edit: I have tried to show it by contradiction. I assumed there exists $\gamma: [0,1] \to ([0,1], [0,1])$ such that $\gamma(0) = (0,0)$ and $\gamma(1) = (1,0)$. Then I noticed that $(\frac{1}{2}, 1]$ is closed in $X$, so $\gamma_1^{-1}((\frac{1}{2}, 1])$ should be closed too. This means that there is a minimal $m$ such that for all $t \in [0,m), \gamma(t) \notin (\frac{1}{2}, 1]$. I don't know if that's the right approach and how to continue.
 A: Your approach is okay. Assume that there exists a continuous $\gamma : [0,1] \to X$ such that $\gamma (0) = (0,0)$ and $\gamma (1) = (0,1)$. Since $C  = (1/2,1] \times \{0\}$ is closed in $X$, the set $S = \gamma^{-1}(C)$ is a closed, hence compact, subset of $[0,1]$. It has a minimum $m \in S$ and we  have $\gamma(m) \in C$. Clearly $m > 0$ because $\gamma(0) \notin C$. Since $U = X \setminus \{(0,0)\}$ is an open neigborhood of $\gamma(m)$ in $X$, there exists $r \in [0,m)$ such that $\gamma([r,m]) \subset U$. The sets $L_n =  \{(x, \frac{x}{n}): x \in (0,1] \}$ are clopen (= closed and open) in $U$ and we have $\gamma(r) \in L_N$ for some $N$. Thus $\gamma^{-1}(L_N)$ is a non-empty clopen subset of $[r,m]$, therefore $\gamma^{-1}(L_N) = [r,m]$, i.e. $\gamma([r,m]) \subset L_N$. This contradicts the fact that $\gamma(m) \in C$.
A similar argument shows that $X$ is not locally path connected. In fact, no open neighborhood of $(0,1)$ can be path-connected because this would imply the existence of a path in $X$ connecting a point in $X \setminus C$ with $(0,1)$.
