How about the following 'conceptual proof'?  The discriminant is defined by the property that, when
$$
p(x) = x^d - s_1 x^{d-1} + \cdots + (-1)^ds_d = (x-t_1)(x-t_2)\cdots(x-t_d)
$$
(i.e., when one substitutes the $i$th elementary symmetric function of the $t_k$ for $s_i$), the discriminant becomes
$$
D(s_1,\ldots,s_d) = \prod_{i<j} (t_i-t_j)^2.
$$
Suppose there were a factorization $D(s_1,\ldots,s_d) = D_1(s_1,\ldots,s_d)D_1(s_1,\ldots,s_d)$ where each $D_i$ had positive degree.  Then by unique factorization, when one substitutes as above, one must be able to write
$$
D_i(s_1,\ldots,s_d) = c_i \prod_{(i,j)\in S_i} (t_i-t_j)
$$
where $c_1$ and $c_2$ are (nonzero) constants and $S_1$ and $S_2$ are disjoint nonempty subsets of the set of pairs $(i,j)$ in $\{1,\ldots,d\}$ whose union is the entire set of distinct pairs in this set.   Thus, for example, $D_1(s_1,\ldots,s_d)$ will vanish when $t_i=t_j$ (for $i\not=j)$ only if $(i,j)$ or $(j,i)$ belongs to $S_1$.  But, of course, one can't detect which of the $t_i$ are equal using only $s_1,\ldots, s_d$, so it follows that $S_1$ must contain either $(i,j)$ or $(j,i)$ for each distinct pair.  The same argument applied to $D_2$ shows that $S_2$ also must contain either $(i,j)$ or $(j,i)$ for each distinct pair.  Thus, one must have
$$
D_i(s_1,\ldots,s_d) = c_i \prod_{i<j} (t_i-t_j)
$$
for some constants $c_1$ and $c_2$, so $D_1/D_2 = c_1/c_2$, so they are constant multiples of each other.  However, one clearly cannot have a polynomial $D_1(s_1,\ldots,s_d)$ such that
$$
D_1(s_1,\ldots,s_d) = c_1 \prod_{i<j} (t_i-t_j),
$$
since the left hand side cannot detect permutations in the $t_i$, whereas the right hand side will change sign when one makes an odd permutation of the $t_i$.

Thus, $D$ is irreducible.