The long ray and the long line are the only non-metrizable 1-manifolds, see e.g. a paper by Peter Nyikos (who also discusses larger dimensions) here (p.2, just after Main Theorem). No proof is given in the above paper (just saying it is easy). Here is a sketch, say the manifold $M$ has an endpoint, and call it "the leftmost point", $a_0$, and then pick a sequence $a_n$ of points "going to the right". If $M=\cup_{n}[a_0,a_n)$ then we are done. Else keep adding points $a_\omega$, $a_{\omega+1}$, etc. Then either we are done at some countable ordinal $\gamma$ (and then we are done, use that for every countable ordinal $\gamma$ there is a subset $T$ of the reals that is order-isomorphic to $\gamma$), or else we define $a_\alpha$ for all $\alpha<\omega_1$, so we get the long ray. It mist be the case that $M=\cup_{\alpha<\omega_1}[a_0,a_\alpha)$ since otherwise we may define $b=\sup{\alpha<\omega_1} a_\alpha$ and our manifold would not be first-countable at $b$, a contradiction. To see that each $\omega<\gamma<\omega_1$ could be thought a subset of $\mathbb R$ fix a bijection $f:\gamma\to \omega$ and any sequence $c_n>0$ with $\sum_n c_n<\infty$, and for each $\beta<\gamma$ define $r_\beta=\sum_{\delta\le\beta}c_{f_\delta}$, then the set $\{r_\beta:\beta<\gamma\}\subset\mathbb R$ is order-isomorphic to $\gamma$.

The long ray and the long line are the only non-metrizable 1-manifolds, see e.g. a paper by Peter Nyikos (who also discusses larger dimensions) here (p.2, just after Main Theorem). No proof is given in the above paper (just saying it is easy). Here is a sketch, say the manifold $M$ has an endpoint, and call it "the leftmost point", $a_0$, and then pick a sequence $a_n$ of points "going to the right". If $M=\cup_{n}[a_0,a_n)$ then we are done. Else keep adding points $a_\omega$, $a_{\omega+1}$, etc. Then either we are done at some countable ordinal $\gamma$ (and then we are done, use that for every countable ordinal $\gamma$ there is a subset $T$ of the reals that is order-isomorphic to $\gamma$), or else we define $a_\beta$ for all $\beta<\omega_1$, so we get the long ray. It mist be the case that $M=\cup_{\beta<\omega_1}[a_0,a_\beta)$ since otherwise we may define $t=\sup_{\beta<\omega_1} a_\beta \in M$ and our manifold would not be first-countable at $t$, a contradiction. To see that each $\omega<\gamma<\omega_1$ could be thought a subset of $\mathbb R$ fix a bijection $f:\gamma\to \omega$ and any sequence $c_n>0$ with $\sum_n c_n<\infty$, and for each $\beta<\gamma$ define $r_\beta=\sum_{\delta\le\beta}c_{f(\delta)}\in\mathbb R$, then the set $\{r_\beta:\beta<\gamma\}\subset\mathbb R$ is order-isomorphic to $\gamma$.

The long ray and the long line are the only non-metrizable 1-manifolds, see e.g. a paper by Peter Nyikos (who also discusses larger dimensions) here (p.2, just after Main Theorem).

The long ray and the long line are the only non-metrizable 1-manifolds, see e.g. a paper by Peter Nyikos (who also discusses larger dimensions) here (p.2, just after Main Theorem). No proof is given in the above paper (just saying it is easy). Here is a sketch, say the manifold $M$ has an endpoint, and call it "the leftmost point", $a_0$, and then pick a sequence $a_n$ of points "going to the right". If $M=\cup_{n}[a_0,a_n)$ then we are done. Else keep adding points $a_\omega$, $a_{\omega+1}$, etc. Then either we are done at some countable ordinal $\gamma$ (and then we are done, use that for every countable ordinal $\gamma$ there is a subset $T$ of the reals that is order-isomorphic to $\gamma$), or else we define $a_\alpha$ for all $\alpha<\omega_1$, so we get the long ray. It mist be the case that $M=\cup_{\alpha<\omega_1}[a_0,a_\alpha)$ since otherwise we may define $b=\sup{\alpha<\omega_1} a_\alpha$ and our manifold would not be first-countable at $b$, a contradiction. To see that each $\omega<\gamma<\omega_1$ could be thought a subset of $\mathbb R$ fix a bijection $f:\gamma\to \omega$ and any sequence $c_n>0$ with $\sum_n c_n<\infty$, and for each $\beta<\gamma$ define $r_\beta=\sum_{\delta\le\beta}c_{f_\delta}$, then the set $\{r_\beta:\beta<\gamma\}\subset\mathbb R$ is order-isomorphic to $\gamma$.