The ordered square $I_0^2$ is connected but not path connected.
My attempt: It’s easy to check $x\times (0,1)=(x\times 0, x\times 1)$, where $x\in I=[0,1]$. Let $f:[a,b]\to I_0^2 $ be a path from $0\times 0$ to $1\times 1$. By intermediate value theorem, $f$ is surjective. Since $x\times (0,1) \neq \emptyset$ and $f$ is surjective, $U_x=f^{-1}(x\times (0,1))\neq \emptyset$ and $U_x \subseteq [a,b]$. Since $f$ is continuous and $x\times (0,1) =(x\times 0, x\times 1)\in \mathcal{T}_{I_0^2}$, $U_x\in \mathcal{T}_{[a,b]}=\mathcal{T}_s =\{ [a,b]\cap V|V\in \mathcal{T}_\Bbb{R} \}$(subspace topology inherits from $\Bbb{R}$). $\forall x\in I$, $\exists z\in U_x$, because $U_x\neq \emptyset$. Since $U_x\in \mathcal{T}_s$ and $z\in U_x$, $\exists (e,f)\in \mathcal{B}_{[a,b]}=\{ B\cap [a,b]| B\in \mathcal{B_{\Bbb{R}}}\}$ such that $z\in (e,f) \subseteq U_x$. Interval $(e,f)$ intersect rational and irrational points. So $\forall x\in I,\exists q_x\in (e,f)\subseteq U_x$ such that $q_x\in \Bbb{Q}$. Munkres skipped this detail of existence of rational number in $U_x$.
Note $U_x\cap U_y=\emptyset$, since $f^{-1}(x\times (0,1)) \cap f^{-1}(y\times (0,1))=f^{-1}([x\times (0,1)] \cap [y\times (0,1)] )=f^{-1}(\emptyset)=\emptyset$. So $q_x \neq q_y$, if $x\neq y$ where $x,y\in [0,1]$. Map $g:I\to \Bbb{Q}$ defined by $g(x)=q_x ,\forall x\in I$ is injective. $g(I)\subset \Bbb{Q}$. Since $g(I)$ is not finite and $\Bbb{Q}$ is countable, $g(I)$ is countable. Since $g$ is injective, $|I|=|g(I)|$. Thus $I=[0,1]$ is countable. Thus we reach contradiction. Can this argument made more precise? because I don’t feel satisfy with my proof.
Best Answer
Basic fact: the lexicographic square $X$ is not separable (but it is compact and thus also Lindelöf; it follows it’s not metrisable).
It’s connected because it’s a linear continuum (which Munkres also shows in the text).
If $f: [0,1] \to X$ is a continuous path from $0\times0$ to $1\times1$ it must indeed be surjective by connectedness and as $[0,1]$ is separable, so is $X$ contradiction. So no path exists.