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Let $a
To prove that $b$ is an upper bound, consider $t$, $t \in T$, since intersection of $Q$ and $[a,b]$ is simply the set $T'=[x \in Q : a \le x \le b]$, therfeore, for every value of $t \in T$, $b \ge t$, hence, $b$ is upper bound of $T$.
To prove second part, let $r$ be some upper bound of $T$, then $t \le r$ for all $t \in T$. Since, $r$ is upper bound of $T$ and $b$ is element of $T$, therefore, $b \le r$. Therefore, $b$ is least upper bound. Proof complete.

I know it is a work of rookie but I am learning. Please suggest if I am wrong somewhere, and also help me out by suggesting precise mathematical language wherever I could have used to make it look really like a proof.

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Your mistake happens at

and $b$ is element of $T$,

because $b$ can be irrational. and then the statement above isn't true (for example, if $b=\sqrt 2$).

You have proven correctly that $b$ is an upper bound (it could be done more quickly by simply saying that if $x\in T$, then $x\in [a,b]$, and if $x\in [a,b]$, then clearly, $x\leq b$, but nevertheless, the proof is OK).

Your proof of least upper bound is perfectly OK if $b$ is rational, but in that case, $b$ is the maximum of $T$ and maximums are always also the supremums of their sets.

So, the work is still ahead of you. You now have to prove that if $b$ is irrational and $r$ is some upper bound of $T$, then $r\geq b$.

To do that, here's a hint:

Try to prove it by contradiction. If $r

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    Why can't I use the fact that $T=\mathbb{Q}\cap[a,b]$ contains only rational numbers between $[a,b]$. Rational intersection real must be rational only? – 2017-01-23
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    @SEMICASA $T$ contains only rational numbers, yes. But $b$ may not be a rational number (and, note, $b$ **may not be** an element of $T$). For example, if $a=0$ and $b=\sqrt 2$, then $b$ **is not an element** of $T$. The task explicitly states *Let $a – 2017-01-23
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    Let there be some $e=b-r>0$. Since we have taken $r$r$ such that $r$b$ is least upper bound. Is this right way to prove contradiction is wrong? – 2017-01-23
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    @SEMICASA Why can't $e<2e$ be true? It's perfectly true for all positive numbers... Again, take a look at my hint. – 2017-01-23
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    yes I was in hurry to prove and made such a mistake. If that is true than contradiction is correct and that should be the case if I am proving $b$ to be least upper bound. – 2017-01-23
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    @SEMICASA No, the proof is nowhere near good. Try again, and *slowly* this time, and please, take a look at my hint. – 2017-01-23
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    Let there be an element $x \in T$, therefore, $x \le b$. Now since $r $r0$, then $r$r$ can not be an upper bound of $T$. I feel happy this time, I think I have got it this time. :-) – 2017-01-23
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    @SEMICASA Again, the proof is not good. Sure, you proved that $r$r$ is not an upper bound of $T$. Again, you didn't listen to my hint. – 2017-01-23
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    How do I find an element in $T$ which is greater than $r$. I understood the hint but I am unable to manipulate it to the desired result. – 2017-01-23
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    @SEMICASA You need to prove that there exists a rational number in $(r,b]$. – 2017-01-23
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    @SEMICASA If you are tired, go rest. I won't solve this any more than I already did, as even a basic look at a textbook should be enough to solve it. – 2017-01-23
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    Supposing the fact that I have shown $r$ is not an upper bound of $T$ for the case $r$r$, $r \ge b$ which I wanted to prove. – 2017-01-24
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    @SEMICASA Because if $r$ is not an upper bound for $r$b$ (because they cannot be smaller). You seem very confused about the subject. I advise you to actually sit down and talk to someone (like your teacher) for a solid session. – 2017-01-24