Since the set $\{1,2\}\times B$ is well ordered, any non-empty subset of this set has a smallest element.
Now consider the subset of $\{1,2\}\times B$ consisting of all those elements whose section is uncountable, clearly, it's non-empty and hence there exists a smallest element, denote it by $\Omega$, hence the result follows!
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 $a$.
Best Answer
The long line is $S_\Omega \times [0,1)$ (not $(0,1]$) in the dictionary order topology (see p. 158 bottom (ex. 12) (all references to the 2nd edition of Munkres))
Then ex. 6 (chapter 3) on that same p. 158 says that if $X$ is well-ordered then $X \times [0,1)$ (in the dictionary order) is a linear continuum. So the long line is a linear continuum.
Finally ex. 8 on p. 206 (chapter 4) says that a linear continuum is normal.
So the long line is normal.
(Side note: it's not too hard but non-trivial, hence Munkres leaves it out of his text book for that reason I think, that any linearly ordered set in the order topology is even monotonically normal, which implies it is hereditarily normal (or completely normal), and this even extends to so-called generalised ordered spaces (which includes the lower limit topology as well). Now the result seems to depend on well-orderedness etc. while in fact it does not, really).
The long line is not metrisable, because it is limit point compact but not compact (for metrisable spaces these notions are equivalent). That's one of the possible arguments for it.