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I tried proving it from the second countable axiom, but couldn't figure out how.

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    $\begingroup$ It's separable, but not second countable. $\endgroup$ – David Mitra May 11 '14 at 16:53
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Suppose that $\mathcal{B}$ is a base for the lower limit topology (the Sorgenfrey line, a.k.a. $\mathbb{S}$). For every $x \in \mathbb{S}$, we know that $[x, x+1)$ is open, contains $x$, so for some $B_x \in \mathcal{B}$ we have that $x \in B_x \subset [x,x+1)$.

If $x \neq y$, say $x < y$, then $y \in B_y$ but $x \notin B_y$ as $B_y \subset [y, y+1)$. So $B_x \neq B_y$.

This shows that $x \rightarrow B_x$ is an injective function from $\mathbb{S}$ into $\mathcal{B}$, so any base has at least size $\mathfrak{c} = |\mathbb{R}|$.

If a space $X$ is metrisable and separable, it has a countable base. The Sorgenfrey line is separable, as $\mathbb{Q}$ is dense in the Sorgenfrey line (every $[a,b)$ contains an open interval $(a,b)$ which always contains points of $\mathbb{Q}$). But as we see that the Sorgenfrey can have no countable base, but is separable, the Sorgenfrey line is not metrisable.

Alternatively, if you have already proved that $\mathbb{S} \times \mathbb{S}$ is not normal (one of the reasons this example is often introduced): if $\mathbb{S}$ were metrisable, so is its square. But metrisable spaces are normal, so contradiction.

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