Show that Hill's Equation $u'' + a(t)u=0$ if $a(t)<0$ for all $t$ then $u\to\infty$ as $t\to\infty$

Your contradiction is that you supposed there is some $t_0>0$ with $u'(t_0) \leq \frac{1}{2}$ and obtained that $u'(t)\geq 1$ for all $t>0$. I don't see how your limit shows that is unbounded (although if you do show that it will get your result.)

Let us synthesize the counterexample.

Substitution $$u' = uv$$ to the Hill equation creates the Riccati equation of $v$: $$v' + v^2 = a(t).\tag1$$ (the earlier example see there).

The homogeneous equation $(1)$ has common solution $$v_c(t) = \dfrac1{t+const},$$ and that gives reason to try the function $v$ in the form of $$v = \dfrac1{t + w(t)},$$ Then $$a(t) = v' + v^2 = \dfrac{-w'(t)}{(w(t) + t)^2}.\tag2$$ The function $a(t)$ must be bounded at $t\to\infty$ and $T$-periodic with $a(t) < 0$.

This can be achieved if you take in account piecewise continuous functions $a$. For example, substitution of the function $$w(t) = \sin\left(\dfrac{3π}{T}\left(t\ \mathrm{mod}\ \dfrac{T}{3} - \dfrac T6\right)\right) + \dfrac{17}{18} T + \dfrac13((t + \dfrac{2T}{3}\mathrm{mod}\ T - t\ \mathrm{mod}\ T)\tag3$$ to $(2)$ (see also Wolfram Alpha graph for $T= 18$) gives the counterexample $a(t)$.