Question: A friend of mine showed me this proof to demonstrate that any algebra variable does not equal any number.

his proof relies on this idea
$$0\neq1$$
$$0x\neq1x$$
$$0\neq1x$$
$$0\neq x$$
Full Proof:
$$\forall n (n \in \Re)$$
$$0\neq1\Rightarrow0(x-n)\neq1(x-n)$$
$$0(x-n)\neq1(x-n)\Rightarrow0\neq x-n$$
$$0\neq x-n\Rightarrow n\neq x$$

I feel like it is a divide by zero fallacy but I'm yet to pinpoint the exact error

Answer: 
The false implication $x \neq y \to ax \neq ay$ is used. This is true iff $a$ is non-zero (in a field), because then the equivalent (by contraposition) $ax = ay \to x=y$ can be shown by division by $a$. So indeed it's a division by $0$ error implicitly.
When we multiply for $(x-n)$ we need to set $x\neq n$ that is precisely the solution we obtain.