Consider $f\colon \mathbb R \to \mathbb R$ given by $$ f(x) = \begin{cases} 1 & x \ge 0\\ 0 & x < 0\end{cases} $$ $f$ is continuous on $\mathbb R -\{0\}$, but cannot be made into a continuous function by change on a set of measure zero.
since there exists dense set of irrational numbers after removing a set of measure zero ,i can show there exists $(x_n),(y_n)$ of irrational number converges to 0, but $(f(x_n)) $ converges to 0 and $(f(y_n))$ converges to 1.Therefore, this function is discontinous at x=0(I didn't add explicit sequence but is this idea of proof correct)
Your proof is, at least, incomplete. You don't explain why, after removing a set with measure $0$, you can always find such a sequence of irrational numbers.
I think that the introduction of irrational numbers in this problem is not the best approach. All you need is to use the fact if $A\subset\mathbb R$ has measure $0$, then $A^\complement$ is dense.