As (implicitly) observed already in Szemerédi's celebrated paper

*Szemerédi, Endre*, **On sets of integers containing no (k) elements in arithmetic progression**, Acta Arith. 27, 199-245 (1975). ZBL0303.10056.

(and perhaps previously), Szemerédi's theorem for a fixed density $0 < \delta_0 < 1$ (such as $\delta_0 = 1/2$), when combined with van der Waerden's theorem, implies Szemerédi's theorem for arbitrary density $\delta > 0$. This is because, once one is given a subset $A$ of integers in $\{1,\dots,N\}$ of density $\delta$, it is not difficult to use the probabilistic method to find $O_{\delta,\delta_0}(1)$ translates of $A$ (by shifts randomly selected between $-N$ and $N$) which cover this interval to density at least $\delta_0$. If one can find a sufficiently long arithmetic progression inside this union of translates, then by van der Waerden's theorem, at least one of these translates also contains a long progression, which gives Szemerédi's theorem for that density $\delta$.

As a consequence of this argument, the gap in difficulty between Szemerédi's theorem for a fixed density $0 < \delta_0 < 1$ (but arbitrary lengths $k$) and for arbitrary densities (and arbitrary lengths) is basically no greater than the difficulty required to prove van der Waerden's theorem (which can be proved in a page or two).

EDIT: the situation is very different if instead one fixes the length $k$ of the progression. As pointed out in comments, Szemerédi's theorem is now easy for very large densities such as $\delta > 1-1/k$, and the difficulty increases as the density lowers (although several proofs of Szemerédi's theorem proceed by a downward induction on density now commonly known as the **density increment argument**). However, in most proofs, the increase in difficulty as $\delta$ decreases is negligible compared to the increase in difficulty as $k$ increases; for instance the $k=3$ case of the theorem, first established by Roth, is substantially easier than the $k>3$ cases. So the van der Waerden reduction given above, which trades the small-$\delta$ difficulty for the large-$k$ difficulty, is generally not useful in practice (in particular, it is largely incompatible with any attempt to induct on $k$, which tends to be a key component of most approaches to this theorem).