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Suppose that $x_1\le 4\lt x_2$.

Observe that $x_1$ will never be equal to $x_2$,

Can I still use $x_1\le x_2$ in any proof, is this always correct?

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    Yes, since $x_1 \leq x_2$ means $$x_1 – 2017-01-15
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    Yes of course. If $A$ is true and $A\implies B$ then $B$ is true. Just remember that $B$ may not be as "strong" as $A.$ That is, $ B$ may fail to imply $A.$ .... If you write all possible "$\implies$" and "$\iff$" in your written proofs you will make fewer errors, and find it easier to find errors..... If you are trying to show that $A\implies C$, it is logical to show that $A\implies B $ and that $ B\implies C. $ – 2017-01-15

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Yes, you can use that notation. The given inequality strictly implies $\boxed{x_1truth of either proposition mentioned in the statement implies truth of the statement. So what you want to write is totally true and correct.

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    Thanks. May I ask the other example I've just added also correct? – 2017-01-15
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    3 things I would like to say: 1) I think the eqn. (1) cannot be obtained from the given 2 equations. You need to correct it. 2)Yes, the eqn. (2) follows from (1) without any problem. 3) I could not understand what you meant by the sentence below the example. [eqn.(1) means "1 star" and eqn (2) means "2 star"]. – 2017-01-15
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    Thanks your feedback, I've fix your 1) issue. – 2017-01-15
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    Well eqn (1) still does not hold. Take $x=1, \delta=5$ and you will get $-1>0$. – 2017-01-15
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    I've deleted the example, because I found that the $\delta\le 4$ has to be asserted correct then with $|x-4|\lt\delta$ to get $x\ge 0$, I just went the wrong direction to solve my problem... – 2017-01-15
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    Yup, thats the case. – 2017-01-15
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If I have one urn with no more than $4$ beans in it, and I have another urn with more than four beans in it, then is it not true that the first urn has no more beans than the second one?