Q- Let $f : N \to Q$ is a bijective function then f is continuous but inverse function $f$ is not continuous at any point of $Q$ .. (Exercise Problem)
my argument -> claim $1^{st}$ - $f$ is continuos - let $c$ be any point of $N$, i.e $|x-c|<\dfrac{1}{2}$ then for any $\epsilon >0$, whenever $|x-c|<\dfrac{1}{2}$ then $|f(x)-f(c)|=|f(c)-f(c)|=0<\epsilon$ so this delta will work for all $c$ belongs to $N$ and it is independent of point as well as epsilon .so $f$ will be uniformly continuos and $f:N\to Q$ is a UC function. Am i correct? I don't need the bijective part to prove the continuity? Since it is given that $f$ is bijective so inverse function exists from $Q \to N$ which is also one-one and onto also, but I can't proceed further. How to show it is not continuous at any point.
Let me rewrite your question in more understandable way:
Proof. "$f$ is continuous" is pretty trivial. $\mathbb{N}$ is discrete so every function $f:\mathbb{N}\to X$ is continuous. For that you just have to realize that every subset $A\subseteq\mathbb{N}$ is open, in particular $f^{-1}(U)$ for open $U\subseteq\mathbb{Q}$.
"$f^{-1}$ is not continuous at any point" Indeed, assume that $f^{-1}$ is continuous at $q\in\mathbb{Q}$ and let $(a_n)\subseteq\mathbb{Q}$ be a sequence convergent to $\mathbb{Q}$. Then $f^{-1}(a_n)$ is a convergent sequence. Since $\mathbb{N}$ is discrete then $f^{-1}(a_n)$ has to be eventually constant. And since $f^{-1}$ is injective then $(a_n)$ has to be eventually constant as well. But we can always find a sequence that is convergent to $q$ that is not eventually constant, namely $a_n=q+\frac{1}{n}$. Contradiction. Hence $f^{-1}$ is not continuous at $q\in\mathbb{Q}$.