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David E Speyer
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The easier part is to see that $\lim_{\infty \leftarrow n} H^*(G(d,n)) \cong \mathrm{Rep}(GL_d)$ as rings. The representation ring of $GL_d$ is the same as the $GL_d$-equivariant $K$-theory of a point. Let $\mathrm{Mat}_{d \times n}^{\circ}$ be the full rank $d \times n$-matrices. I won't define this rigorously, but the ind-scheme $\lim_{n \to \infty} \mathrm{Mat}_{d \times n}^{\circ}$ is "contractible", so the $GL_d$-equivariant $K$-theory of a point is $K^0(\lim_{n \to \infty} \mathrm{Mat}_{d \times n}^{\circ} / GL_d)$. One can justify turning this into $\lim_{n \leftarrow \infty} K^0( \mathrm{Mat}_{d \times n}^{\circ} / GL_d)$. Of course, $\mathrm{Mat}_{d \times n}^{\circ} / GL_d = G(d,n)$.` The Chern character map is an isomorphism between $K^0(G(d,n))$ and $H^*(G(d,n))$, at least once we tensor with $\mathbb{Q}$.

The hard thing is to explain why there should be an isomorphism which takes irreps to Schubert cycles, and works over $\mathbb{Z}$. This is especially difficult because Chern character is not that isomorphism. I don't know any really short answer to this question, but Harry Tamvakis has an expository paper which does a pretty good job.

David E Speyer
  • 156.4k
  • 14
  • 422
  • 763