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I start with analytic geometry approach, i.e. two distinct points $P$=$(x1,y1)$ and $Q$=$(x2,y2)$ with case $x1=x2$ and conclude that the equation of the line is actually $x=x1$ meaning that the value of $x$ is unique. Is my proof true for this case?

Meanwhile, I am stuck with the second possibility, when $x1$ is not the same as $x2$

Please your help

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    What is your definition of a line? You say that you "conclude" that the equation of the line is $x=x_1$, but what is your argumentation leading to this conclusion? – 2017-02-05
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    I mean the line lways gives value $x=x1$ so it is unique – 2017-02-05
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    The line $x=x_1$ is a line that passes through $P$ and $Q$, but you didn't show that there are no other line passing through those points. – 2017-02-05

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If your definition of a line in $\mathbb{R}^2$ is any set of the form $$ L_{m,b}:=\{(x,y)\in\mathbb{R}^2:y=mx+b\} $$ or $$ V_c:=\{(x,y)\in\mathbb{R}^2:x=c\} $$ (vertical lines) then you can proceed like this.

Case 1: Suppose $x_1=x_2$.

Clearly, the only vertical line passing through $P$ and $Q$ is $V_{x_1}$.

Now, no line of the form $L_{m,b}$ passes through $P$ and $Q$. To show this, suppose that $$ y_1=mx_1+b\\ y_2=mx_2+b $$ Then, $$ b=y_1-mx_1=y_2-mx_2\tag{1} $$ so that $$ y_2-y_1=m\underbrace{(x_2-x_1)}_{0}=0\tag{2} $$ and $y_1=y_2$. This contradicts the assumption that $P\neq Q$.

Case 2: Suppose $x_1\neq x_2$.

Clearly no vertical line passes through $P$ and $Q$.

Suppose that $L_{m,b}$ passes through $P$ and $Q$. Then, as in $(1)$, we must have $$ m=\frac{y_2-y_1}{x_2-x_1}\tag{3} $$ and from $(1)$ we must have $$ b=\frac{y_1x_2-y_2x_1}{x_2-x_1}\tag{4} $$ This shows unicity of $m$ and $b$ and since it is readily verified that the line $L_{m,b}$ with $m$ and $b$ as in $(3)$ and $(4)$ passes through $P$ and $Q$, we have existence.

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    in your case $x=x1$, do you mean firstly $y1$ is not the same as $y2$? – 2017-02-05
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    @wawar05 One of your assumption is that the points $P$ and $Q$ are distinct. If we assume that $x_1=x_2$, then you can't have $y_1=y_2$ also because in that case you'd have $P=(x_1,y_1)=(x_2,y_2)=Q$ so $P=Q$ are one and only point. – 2017-02-05
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    I mean, since you used contradiction implyng a contradiction in the last, i.e. $y1$=$y2$, what is your assumption in the first that seemed contradictive to the case or to the theorem? – 2017-02-05
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    @wawar05 If I want to prove that $A$ implies $B$, I can proceed by supposing that $A$ is true and $B$ is false and obtain a contradiction. That contradiction would mean that it is impossible that $B$ is false when $A$ is true, that is, that $B$ must be true when $A$ is true. In this case, $A$ is "$P$ and $Q$ are two distinct points" and $B$ is "there exists a unique line passing through $P$ and $Q$". I split up in two cases, so that in the first case $A$ is really "$P$ and $Q$ are two distinct points such that $x_1=x_2$". Then I supposed that $B$ is false, that is to say that there is... – 2017-02-05
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    ... another line, different from $V_{x_1}$, passing through $P$ and $Q$. I said that clearly such a line must be of the form $L_{m,b}$ (not $V_c$). So I supposed that there was also a line of the form $L_{m,b}$ passing through $P$ and $Q$. Using the hypothesis that $x_1=x_2$, I showed that $y_1=y_2$ must also be true. But that meant that $P=Q$, contradicting the hypothesis in $A$ that $P\neq Q$. Therefore $B$ had to be true, i.e. there was really only one line passing through $P$ and $Q$. – 2017-02-05