Let $X,Y $ be normed spaces. Then a linear transformation $T$ from $X$ to $Y$ is said to be compact if, for any bounded sequence $(x_n)$ in $X$, sequence $T(x_n)$ in $Y$ has a convergent subsequence.
Now I have to show that every compact operator is bounded.
If $T$ is NOT Bounded, then for each $n\in \mathbb{N}$, there exist $x_n\in X$ such that $||T(x_n)||> \ n$.
Then from the definition of compactness I will get a contradiction. Hence it follows that $T$ is bounded.
But my question is how will I show that the sequence $(x_n)$ is a bounded sequence in $X$??
You say "if $T$ is NOT Bounded ,then for each $n\in \mathbb{N}$, there exist $x_n\in X$ such that $||T(x_n)||> \ n$" But this is also true for bounded operators $T$. For instance let $T=I$ be the identity and take $\|x_n\|=n+1$. Clearly $\|T(x_n)\| > n$ but $I$ is a bounded operator.
What you need is that for an unbounded operator $T$, for each $n \in \mathbb{N}$, there exists an $x_n \in X$ with $\|x_n\| =1$ such that $\|T(x_n)\|>n$. From here it becomes clear that a compact operator must be bounded.