In the book Elliptic Carleman Estimates and Applications to Stabilization and Controllability, Volume I page186. The authors proved such a Theorem. In the following $P=\sum_{i,j=1}^{n}\partial_{i}(p_{ij}\partial_{j}\cdot)$ is a second order partial differential operator and the coefficient $p_{ij}$ is smooth. Theorem5.2. Let $\Omega$ be a connected open subset in $\mathbb{R}^{d}$ and let $\omega\subset\Omega$, with $\omega\neq\emptyset$. If $u\in H^{2}(\Omega)$ satisfies$$|Pu(x)|\leq C\Big(|u(x)|+|Du(x)|\Big),\quad,a.e.\quad in\Omega$$ for some $C>0$ and $u(x)=0$ in $\omega$, then $u$ vanishes in $\Omega$.

Sketch of Proof: The authors set $F=supp(u)$(so $F$ is a closed set according the definition of support set of function), and then they prove that $F$ is also open to get $u=0$. They take a point $x^{(1)}$ in $F\setminus \mathrm{int}(F)$ and to get a contradiction by the following Proposition(Proposition5.1 in Lebeau book). Their proof of contradiction is by constructing a family of Ball $\mathscr{B}_{t}=B(x^{(0)},(1-t)r_{1}+tr_{2})$ where $x^{(0)}\in\Omega\setminus F$, and they say according to the following proposition, we can get if u vanishes in $\mathscr{B}_{t}$ with $0\leq t\leq1$, then there exists $\epsilon>0$ such that $u$ vanishes in $\mathscr{B}_{t+\epsilon}$, here, we just know $B(x^{(0)};r_{1})\subset\Omega\setminus F$ in which $u=0$, the question is how can we push the radius from $r_{1}$ to $r_{2}$, the $\epsilon$ in each step is different, does there exist such a situation, that in each step, the increase of radius is$\frac{1}{10},\frac{1}{10^{2}},\frac{1}{10^{3}},\dots$ so that we can not get $B(x^{(0)};r_{2})$? Just in ordinary differential equation, when we study the extension of classical solution of $\dot{x}(t)=f(x(t))$? Everytime we extend a little, but the total step is finite?

Proposition5,1. Let $u\in H^{2}_{\rm loc}(\Omega)$ and $$|Pu(x)|\leq C\Big(|u(x)|+|Du(x)|\Big),\quad,a.e.\quad in\Omega$$ for some $C>0$ and $u(x)=0$ in $\{x\in V;\phi(x)\geq\phi(x^{(0)})\}$. Then $u$ vanishes in a neighborhood of $x^{(0)}$. Remark, the author didn't tell us how large of the neighborhood

In the book Elliptic Carleman Estimates and Applications to Stabilization and Controllability, Volume I, page 186 (MR4436025, Zbl 1497.35005), the authors proves a unique continuation theorem which is the argument of this question. 
In the following $P=\sum_{i,j=1}^{n}\partial_{i}(p_{ij}\partial_{j}\cdot)$ is a second order partial differential operator and the coefficient $p_{ij}$ is smooth.

Theorem 5.2. Let $\Omega$ be a connected open subset in $\mathbb{R}^{d}$ and let $\omega\subset\Omega$, with $\omega\neq\emptyset$. If $u\in H^{2}(\Omega)$ satisfies$$|Pu(x)|\leq C\Big(|u(x)|+|Du(x)|\Big),\quad,a.e.\quad in\Omega$$ for some $C>0$ and $u(x)=0$ in $\omega$, then $u$ vanishes in $\Omega$.

Sketch of Proof: the authors set $F=\operatorname{supp}(u)$ (so $F$ is a closed set according the definition of support set of function), and then they prove that $F$ is also open to get $F=\emptyset$ and thus $u=0$.
They take a point $x^{(1)}$ in $F\setminus \operatorname{int}(F)$ and work in order to get a contradiction by the following proposition (Proposition 5.1 in the book). Precisely, their proof of by contradiction is worked out by constructing a family of balls $$ \mathscr{B}_{t}=B\big(x^{(0)},(1-t)r_{1}+tr_{2}\big) $$ where $x^{(0)}\in\Omega\setminus F$, and then proving (according to the said proposition) that if u vanishes in $\mathscr{B}_{t}$ with $0\leq t\leq1$, then there exists $\epsilon>0$ such that $u$ vanishes in $\mathscr{B}_{t+\epsilon}$.
Here, we just know $B(x^{(0)};r_{1})\subset\Omega\setminus F$ in which $u=0$: now the question is how can we push the radius from $r_{1}$ to $r_{2}$?
The $\epsilon$ in each step is different, so how does there exist such a situation, that in each step, the increase of radius is$\frac{1}{10},\frac{1}{10^{2}},\frac{1}{10^{3}},\dots$ so that we can not intersect $B(x^{(0)};r_{2})$?
Is it just as in ordinary differential equation, when we study the extension of classical solution of $\dot{x}(t)=f(x(t))$? 
Everytime we extend a little, but how the total step is finite?

Proposition 5.1. Let $u\in H^{2}_{\rm loc}(\Omega)$ and $$|Pu(x)|\leq C\Big(|u(x)|+|Du(x)|\Big),\text{ a.e. in }\Omega$$ for some $C>0$ and $u(x)=0$ in $\{x\in V;\phi(x)\geq\phi(x^{(0)})\}$. Then $u$ vanishes in a neighborhood of $x^{(0)}$. 
Remark, the author didn't tell us how large of the neighborhood is.

In the book Elliptic Carleman Estimates and Applications to Stabilization and Controllability, Volume I page186. The authors proved such a Theorem. In the following $P=\sum_{i,j=1}^{n}\partial_{i}(p_{ij}\partial_{j}\cdot)$ is a second order partial differential operator and the coefficient $p_{ij}$ is smooth. Theorem5.2. Let $\Omega$ be a connected open subset in $\mathbb{R}^{d}$ and let $\omega\subset\Omega$, with $\omega\neq\emptyset$. If $u\in H^{2}(\Omega)$ satisfies$$|Pu(x)|\leq C\Big(|u(x)|+|Du(x)|\Big),\quad,a.e.\quad in\Omega$$ for some $C>0$ and $u(x)=0$ in $\omega$, then $u$ vanishes in $\Omega$.

Sketch of Proof: The authors set $F=supp(u)$(so $F$ is a closed set according the definition of support set of function), and then they prove that $F$ is also open to get $u=0$. They take a point $x^{(1)}$ in $F\backslash \mathring{F}$ and to get a contradiction by the following Proposition(Proposition5.1 in Lebeau book). Their proof of contradiction is by constructing a family of Ball $\mathscr{B}_{t}=B(x^{(0)},(1-t)r_{1}+tr_{2})$ where $x^{(0)}\in\Omega\backslash F$,

  and they say according to the following proposition, we can get if u vanishes in $\mathscr{B}_{t}$ with $0\leq t\leq1$, then there exists $\epsilon>0$ such that $u$ vanishes in $\mathscr{B}_{t+\epsilon}$, here, we just know $B(x^{(0)};r_{1})\subset\Omega\backslash F$ in which $u=0$, the question is how can we push the radius from $r_{1}$ to $r_{2}$, the $\epsilon$ in each step is different, does there exist such a situation, that in each step, the increase of radius is$\frac{1}{10},\frac{1}{10^{2}},\frac{1}{10^{3}},\dots$ so that we can not get $B(x^{(0)};r_{2})$? Just in ordinary differential equation, when we study the extension of classical solution of $\dot{x}(t)=f(x(t))$? Everytime we extend a little, but the total step is finite?

Proposition5,1. Let $u\in H^{2}_{\rm loc}(\Omega)$ and $$|Pu(x)|\leq C\Big(|u(x)|+|Du(x)|\Big),\quad,a.e.\quad in\Omega$$ for some $C>0$ and $u(x)=0$ in $\{x\in V;\phi(x)\geq\phi(x^{(0)})\}$. Then $u$ vanishes in a neighborhood of $x^{(0)$. Remark, the author didn't tell us how large of the neighborhood

In the book Elliptic Carleman Estimates and Applications to Stabilization and Controllability, Volume I page186. The authors proved such a Theorem. In the following $P=\sum_{i,j=1}^{n}\partial_{i}(p_{ij}\partial_{j}\cdot)$ is a second order partial differential operator and the coefficient $p_{ij}$ is smooth. Theorem5.2. Let $\Omega$ be a connected open subset in $\mathbb{R}^{d}$ and let $\omega\subset\Omega$, with $\omega\neq\emptyset$. If $u\in H^{2}(\Omega)$ satisfies$$|Pu(x)|\leq C\Big(|u(x)|+|Du(x)|\Big),\quad,a.e.\quad in\Omega$$ for some $C>0$ and $u(x)=0$ in $\omega$, then $u$ vanishes in $\Omega$.

Sketch of Proof: The authors set $F=supp(u)$(so $F$ is a closed set according the definition of support set of function), and then they prove that $F$ is also open to get $u=0$. They take a point $x^{(1)}$ in $F\setminus \mathrm{int}(F)$ and to get a contradiction by the following Proposition(Proposition5.1 in Lebeau book). Their proof of contradiction is by constructing a family of Ball $\mathscr{B}_{t}=B(x^{(0)},(1-t)r_{1}+tr_{2})$ where $x^{(0)}\in\Omega\setminus F$, and they say according to the following proposition, we can get if u vanishes in $\mathscr{B}_{t}$ with $0\leq t\leq1$, then there exists $\epsilon>0$ such that $u$ vanishes in $\mathscr{B}_{t+\epsilon}$, here, we just know $B(x^{(0)};r_{1})\subset\Omega\setminus F$ in which $u=0$, the question is how can we push the radius from $r_{1}$ to $r_{2}$, the $\epsilon$ in each step is different, does there exist such a situation, that in each step, the increase of radius is$\frac{1}{10},\frac{1}{10^{2}},\frac{1}{10^{3}},\dots$ so that we can not get $B(x^{(0)};r_{2})$? Just in ordinary differential equation, when we study the extension of classical solution of $\dot{x}(t)=f(x(t))$? Everytime we extend a little, but the total step is finite?

Proposition5,1. Let $u\in H^{2}_{\rm loc}(\Omega)$ and $$|Pu(x)|\leq C\Big(|u(x)|+|Du(x)|\Big),\quad,a.e.\quad in\Omega$$ for some $C>0$ and $u(x)=0$ in $\{x\in V;\phi(x)\geq\phi(x^{(0)})\}$. Then $u$ vanishes in a neighborhood of $x^{(0)}$. Remark, the author didn't tell us how large of the neighborhood