Moved a comment to an answer as per instructions.

I am not sure I agree that equivariant cohomology is supreme as a candidate for the quickest way to go if one wants an algebro-topological proof. One can use the Eilenberg-Moore spectral sequence which has the form (with a field $k$ as coefficient ring say) $$\mathrm{Tor}^*_{H^*(BK)}(H^*(BT)),k) \implies H^*(K/T)$$ and as $H^*(BT)$ is free as $H^*(BK)$-module (when $k$ has characteristic $0$) this degenerates to give what one wants. In any case if either of these proofs qualifies as "short" depends on whether your definition of length is recursive or not....

No, I don't think so. The intent is certainly that is the same throughout. I do however switch freely back and forth between $G$-representations and $\mathfrak g$-representations as their categories are equivalent.

The map from $C$ to $J(B)$ factors through the map from $C$ to $B$ and hence is almost never an embedding.

Not that I disagree with the conclusion of this answer but the group scheme is $\mathrm{Spec}k[t]\times\mathrm{Spec}k[k]$ with $t\mapsto 1\otimes t+t\otimes1$ and $k[k]$ is the group algebra of the additive group k (and thus $\lambda\mapsto\lambda\otimes\lambda$). (I think that both for this and for the Lie algebra description one has to assume that $k$ is algebraically closed.)

@Kevin: What we need is the other direction, given a dominant weight there is an irreducible representation with that weight as highest weight.

Yes, a line bundle on a compact Kähler variety has a connection iff it has a flat connection. This follows from the Hodge decomposition of de Rham cohomology, the obstruction for finding a flat connection lies in $H^1(X,\Omega^{\ge1})$ and the obstruction for finding a connection is its image in $H^1(X,\Omega^1)$. By the Hodge decomposition (and the fact that the obstruction is of type $(1,1)$), if the latter vanishes so does the former.

(cont'd) Hence, one should rather be looking for a proof that uses as few as possible of the properties of (semi-simple) algebraic groups as the properness of $G/B$ is used to establish such properties.

It seems to me that one of the quickest ways of proving the existence of an irrep with a highest weight vector is to use the properness of $G/B$. I guess it can be done in other ways (by for instance first constructing it over the Lie algebra and integrate). However, I think that this illustrates that maybe this is not such a good question: One wants to have the properness of $G/B$ as early as possible as it has many important consequences (conjugacy of Borel subgroups for one, but also that representations induced from $B$ are finite dimensional).