What is a direct proof for the existence of a weak homotopy equivalence between the Grassmanian $Gr(n):=Gr(n,\infty)$ and the classifying space $BGL(n)$ of $GL(n)$?
They both represent the functor of isomorphism classes of rank $n$ vector bundles on some CW complex but perhaps there is a more direct proof to see this fact.
One possible approach would be the following:
The bar construction gives you a principal $GL_n$ bundle $$GL_n\rightarrow EGL_n\rightarrow BGL(n)$$ with $EGL_n$ being weakly contractible.
For the Grassmannian one has a similar principal bundle $$GL_n\rightarrow V_n(\infty)\rightarrow G_n(\infty),$$ where $V_n(\infty)$ is the Stiefel manifold of $n$-frames and the map $V_n(\infty)\rightarrow Gr_n(\infty)$ is given by taking the span. The classifying map of this bundle gives you a commutative diagram $$\require{AMScd} \begin{CD} GL_n @>{}>> V_n(\infty)@>{}>>G_n(\infty);\\ @VVV @VVV @VVV\\ GL(n) @>{}>> EGL(n)@>{}>>BGL_n; \end{CD}$$ of fiber bundles. The ladder of long exact sequences in homotopy groups then yields the result as $V_n(\infty)$ is weakly contractible.
Sidenote: Maybe it would be cleaner to do this first unstably for the bundle involving $V_n(k)$ and $Gr_n(k)$ to get a highly connected map $Gr_n(k)\rightarrow BGL_n$ and then passing to the colimit.
However, this is not a real answer to your question since I used that $BGL_n$ classifies principal $GL_n$-bundles. All I have proven is that if you have a principal bundle with a weakly contractible total space and you know that there is a classifying bundle, then the base space of the bundle one has started with is weakly equivalent to the base space of the classifying bundle.