Added: The question changed while the answer below was being typed. A short note is added at the end to answer the altered question.
The wording of the original question does not accurately convey the intention. I believe that the OP wants a proof that every real number $x$ is the limit of a sequence $(a_k)$, where each $a_k$ is of the form $m/2^n$, with $m$ an integer and $n$ a non-negative integer. (Numbers of the form $m/2^n$ are called dyadic rationals.)
It is enough to show that for any positive integer $k$, we can find a number $a_k$ of the required form such that $|x-a_k|<1/k$.
We prove the result for $x>0$. A small modification takes care of the case $x<0$.
Let $n$ be the smallest integer such that $2^n>k$. Consider the numbers $0/2^n$, $1/2^n$, $2/2^n$, $3/2^n$, and so on.
Let $m$ be the smallest non-negative integer such that $(m+1)/2^n>x$. Then $m/2^n \le x$. Let $a_k=m/2^n$. Since $\frac{m+1}{2^n} -\frac{m}{2^n}=\frac{1}{2^n} <\frac{1}{k},$ it follows that $0\le x-\frac{m}{2^n}=x-a_k <\frac{1}{k}.$
Added Note: The question was changed and now asks one to show that if $f$ is a continuous function which is $0$ on the dyadic rationals, then $f$ is $0$ everywhere.
Let $x$ be a real number. We show that $f(x)=0$. Let $(a_k)$ be a sequence of dyadic rationals with limit $x$. Such a sequence exists by the calculation above. Then $f(x)=\lim_{k\to \infty} f(a_k) =0.$