There is a theorem that curve $y=f(x)$, where $f:\ [a;b]\rightarrow \mathbb{R}$ and $f$ is a continuous function on this interval, then it's Jordan measure is equal $0$. Why is this so?
I have been taught that Jordan measure is an extension of the notion of size (length, area, volume), so shouldn't the Jordan measure of a curve be it's length? So it should be non-zero for $a\neq b$.
I presume that the author means the Peano-Jordan measure of the graph of $f$ in $[a,b]\times \Bbb R\subseteq \Bbb R^2$.
The graph of a function cannot contain a polyrectangle and therefore its inner Peano-Jordan measure is $0$. This means that if the graph is Peano-Jordan measurable then its measure is $0$.
To see that, in your case, $m^*(\operatorname{gr} f)=0$ as well, notice that $f:[a, b]\to\Bbb R$ is continuous and therefore uniformly continuous. Therefore, for every $\varepsilon$ there is some $\delta_\varepsilon<1$ such that $\lvert f(x)-f(y)\rvert<\varepsilon$ for all $x,y$ such that $\lvert x-y\rvert\le\delta_\varepsilon$. This implies that for all $x\in [a,b]$, $$\left(\left(x-\frac{\delta_\varepsilon}2,x+\frac{\delta_\varepsilon}2\right]\times\Bbb R\right)\cap\operatorname{gr}f\subseteq \left(x-\frac{\delta_\varepsilon}2,x+\frac{\delta_\varepsilon}2\right]\times\left(f(x)-\varepsilon,f(x)+\varepsilon\right]$$
Therefore, for any $\varepsilon>0$ we can select $a=x_0<x_1<\cdots< x_N\le b$ such that $x_n-x_{n-1}=\frac{\delta_\varepsilon}3$ and $b-x_N<\frac{\delta_\varepsilon}3$, and consider the cover $$\operatorname{gr}f\subseteq \bigcup_{j=0}^N \left(x_j-\frac{\delta_\varepsilon}2,x_j+\frac{\delta_\varepsilon}2\right]\times\left(f(x_j)-\varepsilon,f(x_j)+\varepsilon\right]$$
This means that $$m^*(\operatorname{gr}f)\le m^*\left(\bigcup_{j=0}^N \left(x_j-\frac{\delta_\varepsilon}2,x_j+\frac{\delta_\varepsilon}2\right]\times\left(f(x_j)-\varepsilon,f(x_j)+\varepsilon\right]\right)\le \\\le\sum_{j=0}^Nm^*\left(\left(x_j-\frac{\delta_\varepsilon}2,x_j+\frac{\delta_\varepsilon}2\right]\times\left(f(x_j)-\varepsilon,f(x_j)+\varepsilon\right]\right)=2(N+1)\varepsilon\delta_\varepsilon$$
Notice, in addition to that, that by construction $N$ is a very specific number: $$N=\left\lfloor \frac{b-a}{\frac{\delta_\varepsilon}3} \right\rfloor\le 3\frac{b-a}{\delta_\varepsilon}$$
therefore, $$m^*(\operatorname{gr}f)\le (6(b-a)+2\delta_\varepsilon)\varepsilon\le (6(b-a)+2)\varepsilon$$
Taking the $\inf$ over $\varepsilon>0$, we obtain $m^*(\operatorname{gr}f)=0$.