I can't seem to figure out why the proof is false. Everyone knows that $1 \ne 2$ and that $1=2$ is only proven through mathematical fallacies.

However I can't figure out where the problem is in the following fallacy:

Observe that

$$1=\frac{2}{3-1}$$

Substitute $\frac{2}{3-1}$ for the $1$ in the denominator

$$1=\frac{2}{3-\frac{2}{3-1}}$$

Substitute again $$1=\frac{2}{3-\frac{2}{3-\frac{2}{3-1}}}$$

Therefore $1$ can be written as the infinite fraction: $$1=\frac{2}{3-\frac{2}{3-\frac{2}{3-\ldots}}}$$

Now notice that $$2=\frac{2}{3-2}$$

Substitute 2 similar to how 1 was substituted above

$$2=\frac{2}{3-\frac{2}{3-2}}$$

And through repeated substitution we get $$2=\frac{2}{3-\frac{2}{3-\frac{2}{3-\ldots}}}$$

Notice that $1=\frac{2}{3-\frac{2}{3-\frac{2}{3-\ldots}}}$ and $2=\frac{2}{3-\frac{2}{3-\frac{2}{3-\ldots}}}$

Therefore, $1=2$

Just looking at this I know that it cannot be true. But they both seem to share the same infinite fraction, so what is it that I am missing?