I want to prove that
$$(x-1)(x-2)=0\Leftrightarrow x=1, 2$$
$\Leftarrow$ is easy. The problem is $\Rightarrow$.
Assuming $x\neq 1, 2$, we can derive $1=0$ by dividing both sides of $(x-1)(x-2)=0$ by $x-1$ and $x-2$.
Thus we get $\lnot \lnot (x=1, 2)$. However, intuitionistic logic cannot eliminate double negation.
On
The second implication would follow if $\neg (x=1 \wedge x=2)$. This is equivalent to $(x-1)\neq 0 \lor (x-2)\neq 0$, and so: $$ \neg (x=1\wedge x=2) \wedge (x-1)(x-2)=0 $$ is equivalent to: $$((x-1)\neq 0 \wedge (x-1)(x-2)=0)\lor((x-2)\neq 0 \wedge (x-1)(x-2)=0)$$ which is equivalent to: $$((x-1)\neq 0 \wedge (x-2)=0)\lor((x-2)\neq 0 \wedge (x-1)=0)$$ In the first case, with the right and you get $x=2$, in the second you get $x=1$.
I do think the first claim can be proven constructively ($x=1 \wedge x=2$ implies $1=2$ which leads to contradiction, hence the negation holds). But please do correct me if i'm wrong (i'm fairly new to constructive thinking).
On
Real numbers have a property called locatedness, which states that for all $a, b, c$ with $a < b$, we have $a < c \lor c < b$. Suppose $(x - 1)(x - 2) = 0$. Apply locatedness to $a = 1$ and $b = 2$ to conclude that either $1 < x$ or $x < 2$.
If $1 < x$, then $0 < x - 1$, so $x - 1$ has a multiplicative inverse. Therefore, $x - 2 = 0$, so $x = 2$.
If $x < 2$, then $x - 2 < 0$, so $x - 2$ has a multiplicative inverse. Therefore, $x - 1 = 0$, so $x = 1$.
Thus, either $x = 1$ or $x = 2$.
A useful construction on real numbers and more generally is that of a apartness relation. Such a relation is defined to be antireflexive, symmetric and cotransitive, meaning that if $x \# z$ then $x \# y$ or $y \# z$. Classically, this is just the negation of an equivalence relation. Additionally, a tight apartness relation satisfies $\neg (x\ \#\ y) \Rightarrow x = y$.
As you'll see proved in any book on constructive analysis, the real numbers have such a relation given by $x$ and $y$ are apart if there is a rational number between them, i.e., $\exists q \in \mathbb{Q}\ (x < q \land q < y) \lor (y < q \land q < x)$. You'll also find that if $x \# 0$, we can divide by $x$.
Now, given $x\ \#\ y$, we can prove that $x y = 0$ implies $x = 0$ or $y = 0$. This is fairly simple. Since $x\ \#\ y$, cotransitivity tells us that $x\ \#\ 0$ or $y\ \#\ 0$. In the first case, since we can divide by $x$, $x y = 0 = x \cdot 0$ implies that $y = 0$. Similarly, in the second case $x = 0$, so $x = 0$ or $y = 0$.
To apply this to your situation, we need to prove that $(x - 1)\ \#\ (x - 2)$. This seems pretty plausible, since the two numbers are pretty far apart. In fact, any good enough rational approximation of $x - 3/2$ will do.
To procure such a rational approximation, you need some details about which reals you're using. For Cauchy reals, this is trivial since a Cauchy real is equipped with a function that gives a rational approximation to that number with any given precision. We just need to take a rational $q$ within $1/2$ of $x - 3/2$ and that rational will be between $x - 2$ and $x - 1$. $x - 3/2 - q < 1/2$ implies $x - 2 < q$ and $q - (x - 3/2) < 1/2$ implies $q < x - 1$.
For Dedekind reals, we have to be a bit more creative. To start, a Dedekind cut is given by two sets $L$ and $R$ of rationals. They must both be inhabited (there exists an element of each). $L$ should be lower rounded, meaning that $q$ is in $L$ if and only if there is another rational $r$ in $L$ with $q < r$. Similarly, $R$ must be upper rounded. The cuts must be ordered: if $q$ is in $L$ and $r$ is in $R$, the $q < r$. Finally, the cuts must be located: if $q < r$, then either $q$ is in $L$ or $r$ is in $R$.
To get arbitrarily close rational approximations to a given Dedekind cut, we can use inhabitedness to get a starting point and locatedness to get a better rational approximation.
Let $x = (L, R)$ be a Dedekind cut and let $\epsilon > 0$ be some rational number. We want to find an interval with rational endpoints that both contains $x$ and has length less than $\epsilon$.
By inhabitedness, there exists $q_0$ in $L$ and $r_0$ in $R$. By the ordered property, $q_0 < r_0$, so $r_0 - q_0 > 0$. Let $N$ be positive integer such that $(r_0 - q_0)/N < \epsilon$.
Then divide $(q_0, r_0)$ into $N$ subintervals. Explicitly, let $q_k = q_0 + k \cdot (r_0 - q_0)/N$.
By the locatedness property, either $q_k$ is in $L$ or $q_{k + 1}$ is in $R$. Since $q_0$ is in $L$ and $q_N = r_0$ is in $R$, not all of the $q_k$ are in one set. Furthermore, if $q_k$ is in $L$, then $q_i$ is in $L$ too for $i < k$ and vice versa for $R$. This means that there is a unique $n$ such that $q_n$ is in $L$ and $q_{n + 1}$ is in $R$. Thus, $(q_n, q_{n + 1})$ is an interval of length $(r_0 - q_0)/N < \epsilon$ that contains $x$. Both $q_n$ or $q_{n + 1}$ are rational approximations within $\epsilon$ of $x$.
To apply this to our problem, we needed a rational strictly between $x - 1$ and $x - 2$. To get this, use a rational approximation of $x - 3/2$ that is accurate to (strictly) within $\epsilon = 1/2$.