It is very elementary to show that R isn’t homeomorphic to Rm for m>1: subtract a point and use the fact that connectedness is a homeomorphism invariant.
Along similar lines, you can show that R2 isn’t homeomorphic to Rm for m>2 by subtracting a point and checking if the resulting space is simply connected. Still straightforward, but a good deal less elementary.
However, the general result that Rn isn’t homeomorphic to Rm for n≠m, though intuitively obvious, is usually proved using sophisticated results from algebraic topology, such as invariance of domain or extensions of the Jordan curve theorem.
Is there a more elementary proof of this fact? If not, is there intuition for why a proof is so difficult?
There are reasonably accessible proofs that are purely general topology. First one needs to show Brouwer’s fixed point theorem (which has an elementary proof, using barycentric subdivion and Sperner’s lemma), or some result of similar hardness. Then one defines a topological dimension function (there are 3 that all coincide for separable metric spaces, dim (covering dimension), ind (small inductive dimension), Ind (large inductive dimension)), say we use dim, and then we show (using Brouwer) that dim(Rn)=n for all n. As homeomorphic spaces have the same dimension (which is quite clear from the definition), this gives the result. This is in essence the approach Brouwer himself took, but he used a now obsolete dimension function called Dimensionsgrad in his paper, which does coincide with dim etc. for locally compact, locally connected separable metric spaces. Lebesgue proposed the covering dimension, but had a false proof for dim(Rn)=n, which Brouwer corrected.
One can find such proofs in Engelking (general topology), Nagata (dimension theory), or nicely condensed in van Mill’s books on infinite dimensional topology. These proofs do not use homology, homotopy etc., although one could say that the Brouwer proof of his fixed point theorem (via barycentric division etc.) was a precursor to such ideas.