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Proposition: Suppose X is a metric space, and also a topological space with the topology given by the metric. Suppose E is a subset of X. Then the induced metric on E gives the same topology on E as the topology induced from the topology on X.

I encountered the above proposition in a book and it seemed very elementary. But I do not quite know how to prove it. Can I ask for someone's help? Maybe firstly explain the statement a little bit since it is not quite clear to me. Thanks so much.

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    Compare the induced topology from $X$ to $E$ to the topology of the space $E$ induced by the metric, that is, observe if an open set of one topology is an open set in the other topology and viceversa. – 2017-02-20
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    I am sorry I still can't do that by myself. Can you please be a little bit more specific? – 2017-02-20
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    Can you just edit and type that proposition? Image is taking too much time to load. – 2017-02-20
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    @ButterMath how are defined the open sets in $E$ induced from the topology on $X$? How are defined the open sets in $E$ induced by the metric? Compare both cases and you will see that they are equivalent knowing that the topology in $X$ is induced by it metric (how are the open sets of $X$ defined?) – 2017-02-20
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    @VikrantDesai I have edited the question with the proposition typed. – 2017-02-20

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