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Does there exists a function defined on $[0,1] rightarrow mathbb{R}$ such that it contains only jump discontinuity but it is not piecewise continuous?

Jump discontinuity at a point $a$ means $lim_{x rightarrow a^{-}}f(x)$ and
$lim_{x rightarrow a^{+}}f(x)$ both exists but not equal

piecewise continuous means every finite subinterval only contains a finite number of discontinuous points and they are all jump discontinuity

My first thought is Dirichlet function and but it appears that it is not the function that I am looking for.....

OK, next thought - the function $f(frac pq)=frac1q$ and zero elsewhere. That's closer; it has limits of zero everywhere. But then each rational is a removable discontinuity, not a jump discontinuity. Closer, but still not it.

The next idea after that: let's build an increasing function with jumps at every rational. Let $g$ be an enumeration of the rationals; for each rational $r$, $g(r)$ is a different positive integer $n$. Then, define
$$f(x) = sum_{rinmathbb{Q},rle x}frac1{g(r)^2+g(r)}$$
Since $sum_n frac1{n^2+n}$ converges (to $1$), that sum is finite for every $x$.

Choose some arbitrary $x$ and $epsilon>0$. Let $n$ be such that $epsilongefrac1n$. There are only finitely many values $r_1,r_2,dots,r_n$ with $g(r_i)le n$. If we choose $delta$ such that $(x,x+delta)$ contains none of these $r_i$, then for $yin (x,x+delta)$,
$$f(y)-f(x)=sum_{rinmathbb{Q},x<rle y}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
From that, $lim_{yto x^+}f(y)=f(x)$ for all $x$. We have limits from the right.

For limits from the left, consider the variant function
$$f^*(x)=sum_{rinmathbb{Q},r< x}frac1{g(r)^2+g(r)}$$
This $f^*$ is equal to $f$ except at the rationals, where $f(r)-f^*(r)=frac1{g(r)^2+g(r)}$. Again, choose arbitrary $x$ and $epsilon>0$, and let $n$ be such that $epsilonge frac1n$. Find $delta$ such that $(x-delta,x)$ contains none of the $n$ points $r_i$ with $g(r_i)le n$. Then, for $yin (x-delta,x)$,
$$f^*(x)-f(y) = sum_{rinmathbb{Q},yle r< x}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
From that, $lim_{yto x^-}f(y) = f^*(x)$ for all $x$, and we have limits from the left.

Of course, these limits $lim_{yto x^+}f(y)=f(x)$ and $lim_{yto x^-}f(y) = f^*(x)$ differ for every rational $x$, so there's a jump discontinuity at every rational.

With $f$ discontinuous at a dense set of points, it fails to be continuous on any interval, and can't be a piecewise continuous function. Done. We have our example.

I defined this as a function from $mathbb{R}$ to $mathbb{R}$, but it's easy to get a function on a smaller interval. Restricting $f$ works, as does using an enumeration of the rationals in that smaller interval.

$$f(x)=sum_{k=1}^infty 2^{-k} ( 2^kx-lfloor 2^kxrfloor)$$ has a jump discontinuity at every $frac{n}{2^k}$ and it is continuous everywhere else

$$g(x)=sum_{k=1}^infty 2^{-k} ( lfloor 2^kxrfloor-2lfloor 2^{k+1}xrfloor)$$ is easier to see : if $x in [frac{N}{2^k},frac{N+1}{2^k}]$ then $g(x) = frac{N}{2^k}+ O(2^{-k})$