We know that if $A$ is a separable $C^{*}$-algebra then $K_1(A)$ is countable.
Can anybody give an example of a C*-algebra for which $K_1(A)$ is uncountable?
There must be tons of ways to do this, but a simple one is to start with an uncountable set $X$, equipped with the discrete topology, and consider $c_0(X)$. There are uncountably many pairwise inequivalent minimal projections in this algebra, so its $K_0$ group is uncountable. Now use $K_0(c_0(X)) \cong K_1(Sc_0(X))$ where $SA$ is the suspension of $A$.
Nik's answer nails it but if you prefer something representable on a separable Hilbert space then you may consider the suspension $SM$ of any ${\rm II}_1$-factor $M$. Indeed, as $M$ is tracial, $K_0(M) \cong \mathbb R$ and the suspension simply reverses the $K$-groups.
Actually you may produce further commutative examples that are representable on separable Hilbert spaces: for example $S\ell_\infty$, since $K_0(\ell_\infty)$ comprises $\mathbb{Z}$-valued continuous functions on $\beta \mathbb N$.