Fundamental Group of the Long Line

The key to understanding questions like this about the long line is the following observation.

Lemma: For any $x\in L$, the interval $I_x=\{y\in L:y\leq x\}$ is order-isomorphic to $[0,1]$.

Proof: Let $A$ be the set of $y\in L$ such that $(0,0)<y<x$ and the second coordinate of $y$ is rational. Then $A$ is countable (since the first coordinate of $x$ is a countable ordinal so there are only countably many choices for the first coordinate of $y$) and densely ordered without endpoints, so it is order-isomorphic to $(0,1)\cap\mathbb{Q}$. Also, $A$ is dense in $I_x$ and $I_x$ is Dedekind-complete (and has endpoints), so $I_x$ is the Dedekind-completion of $A$ (including endpoints). So, $I_x$ is order-isomorphic to the Dedekind completion (including endpoints) of $(0,1)\cap\mathbb{Q}$, which is just $[0,1]$. $\blacksquare$

This Lemma is really the reason that $L$ is called the "long line". If you cut it off at any point before the end, it really does just look like the ordinary real line. To see that it is different, you have to go all the way "uncountably far" to the end.

In particular, it follows that $I_x$ is homeomorphic to $[0,1]$ and contractible. But any continuous map from a separable space to $L$ has image contained in $I_x$ for some $x\in L$ (take $x$ to be an upper bound for the image of a countable dense subspace). So, any continuous map from a separable space to $L$ is nullhomotopic. In particular, this implies all the homotopy groups of $L$ are trivial.

(Note that the Lemma also implies that the image of a loop in $L$ does not have to be covered by finitely many sets of the form $\{a\}\times[0,1)$. Indeed, for any infinite ordinal $\alpha<\omega_1$, the interval $I_{(\alpha,0)}$ contains infinitely many sets of the form $\{a\}\times[0,1)$, but can be covered by a single loop since it is homeomorphic to $[0,1]$.)

Yes, it suffices to show that the image of any continuous map $f:S^1\rightarrow L$ is bounded, i.e., contained in $\alpha\times[0,1)$ for some countable $\alpha$, and then use that $\alpha\times[0,1)\simeq[0,1)$. This is because $L$ has the so-called $\omega$-compactness property: the closure of any countable set is compact, and also $S^1$ is separable (actually even compact as pointed out in the comment).