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Consider a two-point set $M = \{a, b\}$ whose topology consists of the two sets, $M$ and the empty set. Why does this topology not arise from a metric on $M$?

I'm not sure what this question is asking me to do? Am I to show that $M$ is not metrizable?

I am sorry if I am not providing an attempt at an answer, I'm just completely lost here.

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    Yes, I think you should prove that $M$ is not metrizable. – 2017-02-05
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    Suppose you put a metric on $M$. What will your open sets look like? – 2017-02-05
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    There wouldn't be any open sets? I cant think of how they could exist with a finite set of points like this – 2017-02-05

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Assume $d$ is a metric on the set $M = \{a, b\}$ where $a \neq b$ and let $\delta = d(a, b)$. Then $\delta \neq 0$ because otherwise $a = b$. The set $U = \{x \in M \mid d(x, a) < \delta\}$ is then open by definition of the topology on a metric space. But $U = \{a\}$ is not open in the topology described in the question (the indiscrete topology). By an extension of this argument (taking $\delta = \min\{d(x, y) \mid x, y \in M, x\neq y\}$) you can show that a finite metric space has the discrete topology: the topology in which every subset is open.

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    I am not too familiar with indiscrete topology so I'm not sure how we can know that the described topology is indiscrete. Is it because the only open sets are M and the empty set? – 2017-02-05
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    That is the definition of the indiscrete topology. You don't need to know that bit of terminology to answer the question. – 2017-02-05
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Any topology arising from a metric is Hausdorff. But with $\{\emptyset,M\} $ you cannot separate $a $ and $b $.

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    We havent learned about Hausdorff spaces yet, just metric spaces, is there a similar solution that doesn't depend on Hausdorff? Thanks – 2017-02-05