I am reading Munkres' Topology. Munkres gave two examples of spaces that are not path connected: the ordered square (A.K.A. the unit square in lexicographic order topology) and topologist's sine curve. It seems these two cases are intuitively similar to each other, but Munkres used different techniques to prove them. So I tried to mimic Munkres' proof in the sine curve case to prove the path-disconnectedness for the ordered square.
I tried to proceed as follows. Suppose there is a path $f: [a,c] \to I_o^2$ joining the two points $0 \times 0$ and $1 \times 1$. The set $A = 0 \times [0,1]$ is closed in $I_o^2$, so is $f^{-1}(A)$ in $[a,b]$. So the latter has a largest element $b$. Replace $f$ with its restriction to $[b,c]$; without a loss of generality we also replace $[b,c]$ with $[0,1]$. Then $f$ maps $0$ to some point in $0\times[0,1]$, and any point from $(0,1]$ to some point in $x\times[0,1]$ with some $x\gt 0$.
Then I tried to deduce a contradiction from the continuity of $f$ at the point $0$. It went easy assuming the case $f(0) = 0\times y$ with $y\lt 1$, since one could always find a neighbourhood of $0\times y$ which is completely contained in $0\times[0,1]$. But I could not solve the case where $f(0) = 0 \times 1$. Can we proceed in this way? Or maybe I went into a completely wrong direction?
It's always hard to say conclusively that an approach cannot work, but I suspect that's so in this case.
All proofs that the topologist's sine curve $\bar{S}$ isn't path connected, are related to this fact: $\bar{S}$ isn't locally connected.
(Or at least all the proofs I've seen. Some use this fact more or less explicitly. I'd say Munkres' proof uses the reason that $\bar{S}$ isn't locally connected: A small open disk around $(0,y_0)$ will restrict $x$ to an interval $0\leq x<\epsilon$, but the graph of $\sin(1/x)$ in that interval takes on all $y$-values between $-1$ and $1$. Hence the intersection of the disk with $S$ is not connected.)
(Compare with this space: $\bar{T}=\{(0,0)\}\cup T$, where $T$ is the graph of $x\sin(1/x)$. Since $\lim_{x\to 0}x\sin(1/x)=0$, this space is itself a path, i.e., a continuous image of the unit interval.)
The ordered square, on the other hand, is locally connected. (Munkres gives this as an exercise.) His proof of its non-path-connectedness appeals instead to the fact that $I_o^2$ isn't second-countable.
The Hahn-Mazurkiewicz Theorem is relevant here. It states:
If a topological space $X$ contains a path $f:I\to X$, then the image $f(I)$ with the subspace topology will enjoy all these properties. For the topologist's sine curve, it's local connectivity that fails (for a suitable path); for $I_o^2$, second-countability.