The usual construction for finding torsion elements on complex $K$ theory is using flat vector bundles. So is it still possible to find a simply connected compact space with a nonzero torsion in its $K$ theory. Such an example would be given by a map $f:X\to BU$ which is zero on real cohomology but is not null homotopic. By Bott periodicity theorem, it follows that such a map would be zero on homotopy groups (because $\pi^s_*(Y)\otimes \mathbb{R}=H_*(Y, \mathbb{R})=H^*(Y,\mathbb{R})$ and the homotopy groups of $BU$ are free abelian). I don't know if such a map could exist or not. I am most interested in the case of a closed simply connected manifold but an example in the general case would be welcome.
Let $X$ be a $2$connected closed $7$manifold with $H_3(X) = H^4(X) = \mathbb{Z}/2$. Then $\tilde K(X) = \widetilde{KU}^0(X)$ equals $\mathbb{Z}/2$. This follows from the AtiyahHirzebruch spectral sequence.
If you do not require $X$ to be a manifold, then the $4$skeleton of the manifold above, i.e., $S^3 \cup_2 e^4 = \Sigma^2 \mathbb{R} P^2$, is a slightly smaller example.

$\begingroup$ Thank you for your answer but I don't see why such a manifold exists? $\endgroup$ – Omar Oct 16 '16 at 10:41

3$\begingroup$ @Omar: let $P_k$ the principal $S^3$ bundle over $S^4$ corresponding to the integer $k\in \pi_4(BS^3)\cong \pi_3(S^3)$. Then $P_k$ is $2$connected and $H_3(P_k)$ has order $k$, see e.g. arxiv.org/abs/math/0007198, proposition 3.3. Take $k=2$. $\endgroup$ – Igor Belegradek Oct 16 '16 at 11:49
