I am trying to prove the following claim: The flat $2$-dimensional torus cannot be isometrically immersed into $\mathbb{R}^2$ with the standard metric.
That is, there is no immersion $f:T^2 \rightarrow \mathbb{R}^2 $ which is also an isometry.
Remark: There is no isometric embedding of the flat torus into the Euclidean plane.
Proof:
Any such embedding will be in particular an embedding into $\mathbb{R}^3$. The torus is compact, so its embedded image in $\mathbb{R}^3$ will be also compact, hence will include an elliptic point.
This contradicts the fact that isometric embeddings preserves curvature.
Best Answer
Since there is some ambiguity concerning what are your hypotheses, it is not guaranteed that your proof is exhaustive. It is known that there is a $C^1$-isometric embedding of the flat torus inside $\mathbb{R}^3$ (this is related to Nash embedding theorem). However, for the reason you presented and because of Gauss' theorema egregium, there is no $C^2$-isometric embedding of the flat torus inside $\mathbb{R}^3$.
In fact, there is no immersion $f$ of any closed (i.e compact and without boundary) smooth surface $\Sigma$ into $\mathbb{R}^2$. Indeed, any such immersion would be a fortiori continuous and thus $f(\Sigma)$ would be compact. In particular, there would exist a boundary point $y \in \partial f(\Sigma)$ and for any $x \in f^{-1}(y)$, the differential map $df_{x} : T_x \Sigma \to T_y\mathbb{R}^2$ would not have maximal rank 2 (for otherwise, by the inverse function theorem, $f$ would be a local diffeomorphism between an open neighborhood of $x$ and an open neighborhood of $y$ in the plane, which is clearly not the case). So $df_x$ is not an immersion after all.