Showing that the loopspace $\Omega S^{\infty}$ is homotopic to $S^{\infty}$. Showing that the infinite dimensional sphere $S^{\infty}$ is contractible is rather easy by constructing an explicit contraction (Hatcher gives a nice one). I thought it might be a nice exercise to try and show this using loop spaces and the fact that $\pi_1(S^{\infty})$ is trivial.

Let the loopspace $\Omega X$ be the space of base pointed loops in $X$. For now, assume that the loopspace $ΩS^{\infty}$ is homotopy equivalent to $S^{\infty}$. Then, $\pi_2(S^{\infty})=\pi_1(\Omega S^{\infty})=\pi_1(S^{\infty})=0$. Continuing by induction, we get that every homotopy group of $S^{\infty}$ is trivial and so $S^{\infty}$ is contractible because $S^{\infty}$ is a CW-complex (by Whitehead). As far as I can tell, this argument has no holes in it.

My problem now is in proving the assumption that $\Omega S$ is homotopy equivlanet to $S$. I haven't had too much experience with loopspaces or function spaces in general so I'm struggling somewhat.

One approach might be to find a fibration $F\rightarrow\Omega S^{\infty}\rightarrow S^{\infty}$ and then playing about with long exact sequences in homotopy, but I'm not sure if such a fibration exists. Maybe something like 'If $f\colon [0,1]\rightarrow S^{\infty}$ is a loop, let $p\colon \Omega S^{\infty}\rightarrow S^{\infty}$ be given by $p(f)=f(1/2)$' would work? Is p a fibration? And if so, what do the fibers look like? The best case scenario would be for some fibration to exist with contractible fiber. However, I'm worried that showing such a fiber is contractible would amount to showing that $S^{\infty}$ is contractible, which defeats the object of the exercise.
You might be able to get something using the Freudenthal theorem (though this is a bit overkill admittedly), since it studies the connectedness of the natural map to $\Omega \circ S$. Supposing the assertion that the suspension of $S^\infty$ is again $S^\infty$ (we have to use the limiting behavior of this space somewhere), then if $S^\infty$ is $n$-connected, the map

$\pi_k(S^\infty) \to \pi_k(\Omega S^\infty) = \pi_{k+1}(S^\infty)$

is an isomorphism for $k \le 2n$. Then this is certainly true for $k = n$, which gives that $\pi_{n+1}(S^\infty)$ is trivial as well, and so $S^\infty$ is at least $(n+1)$-connected -- we can continue on to hit all the homotopy groups.
Since $\pi_1(S^\infty)=0$, it is rather easy to prove that $S^\infty$ and $\Omega S^\infty$ are homotopy equivalent : set $\rho : S^\infty \to \Omega S^\infty$, $\rho(p)(t)=p$ for every $t\in[0,1]$ (since $\rho(p)$ is a path, it makes sense to write $\rho(p)(t)=p$) and $\tau : \Omega S^\infty \to S^\infty$, $\tau(\gamma)=\gamma(0)$. Now $\tau\circ\rho$ is the identity, and $\rho\circ\tau$ is isotopic to the identity, since $\pi_1(S^\infty)=0$, and so every loop is homotopy equivalent to the constant loop on its base point : choose an isotopy $F_\gamma$ from each loop $\gamma$ to its base point, then set $F:\Omega S^\infty \times [0,1]\to\Omega S^\infty$ by $F(\gamma,t)=F_\gamma(t)$, and this is a homotopy between the identity and $\rho\circ\tau$.