Can the long line be embedded in the ordered plane?
For $\mathbb{R}^2$ with the lexicographic order, this is very simple, Note that for each $x\in\mathbb{R}$, $\{x\}\times\mathbb{R}$ is clopen in $\mathbb{R}^2$ and homeomorphic to $\mathbb{R}$ with the usual topology. Since the long line is connected, any continuous map from the long line to $\mathbb{R}^2$ must have its image contained in $\{x\}\times\mathbb{R}$ for some $x$, and thus cannot be an embedding since the long line does not embed in $\mathbb{R}$.
For the unit square the argument is a bit more complicated. Let $L$ denote the long line and let $S$ denote $[0,1]^2$ with the lexicographic order. Let $p:S\to [0,1]$ be given by $p(x,y)=x$; note that $p$ is continuous (for the usual topology on $[0,1]$). So, if $f:L\to S$ is any continuous map, we have a continuous composition $p\circ f:L\to [0,1]$. Any continuous map $L\to[0,1]$ is eventually constant, so $p\circ f$ is eventually constant. However, this means that there exists $a\in L$ and $x\in [0,1]$ such that $f(b)\in \{x\}\times[0,1]$ for all $b>a$. Since $\{x\}\times[0,1]$ just has the usual topology of $[0,1]$, this again implies that $f$ is eventually constant, and so cannot be an embedding.
(This second argument applies equally well to the space $\omega_1$ instead of $L$, to show that it does not embed in either $\mathbb{R}^2$ or $[0,1]^2$ with the lexicographic order topology.)
Just for completeness, here is a proof that any continuous map $f:L\to\mathbb{R}$ is eventually constant (the same argument also applies with $\omega_1$ in place of $L$). First, fix $\epsilon>0$. Suppose that for all $a\in L$ there exist $b,c>a$ such that $|f(b)-f(c)|>\epsilon$. We can then choose an increasing sequence $b_1<c_1<b_2<c_2<\dots$ such that $|f(b_n)-f(c_n)|>\epsilon$ for each $n$. Every sequence in $L$ is bounded above, so this sequence has a supremum $x$ which it converges to. But this contradicts the continuity of $f$, since the sequence $(f(b_1),f(c_1),f(b_2),f(c_2),\dots)$ is not Cauchy and so cannot converge to $f(x)$.
So for every $\epsilon>0$, there exists $a\in L$ such that $|f(b)-f(c)|<\epsilon$ for all $b,c>a$. For each $n$, choose $a_n$ that works as such an $a$ for $\epsilon=1/n$. Letting $a$ be an upper bound for all these $a_n$, we have $|f(b)-f(c)|<1/n$ for all $n$ if $b,c>a$. That is, $f(b)=f(c)$ for all $b,c>a$, so $f$ is constant above $a$.