How to find I(t)? I'm working with a SIS model for diseases. Where S stands for susceptibles, and I stands for infected. 
I have a situation that is modeled by the system:
$$S'(t)=\frac{dS}{dt}=-\beta SI-\lambda S$$
$$I'(t)=\frac{dI}{dt}=\beta SI-\alpha I$$
Show that both S(t) and I(t) approach zero as $t \rightarrow \infty$
K is the unit that population size is measured in so we can say that $S=K-I$ and that $I=K-S$. 
These equations look like separable equations to me. Once I find S(t) and I(t) I can easily find the limit.

If I am understanding you properly, we have:
$$S'(t)=\frac{dS}{dt}=-\beta SI-\lambda S$$
$$I'(t)=\frac{dI}{dt}=\beta SI-\alpha I$$
$K$ is the unit that population size is measured in, so we can say that $S=K-I$ and that $I=K-S$.
For the first equation, we have:
$$\frac{dS}{dt}=-\beta SI-\lambda S = -\beta S(K-S)-\lambda S$$
For the second equation, we have:
$$\frac{dI}{dt}=\beta SI-\alpha I = \beta (K-I)I-\alpha I$$
So, now you have them decoupled, and assuming all other variables are constants, you can solve each one separately using Separation of Variables.
This yields:
$$S(t) = -\frac{\beta K+\lambda}{e^{\left(c_1+t\right) (\beta K+\lambda)}-\beta}$$
$$I(t)=\frac{(\beta K-\alpha) e^{\beta K \left(c_1+t\right)}}{\beta e^{\beta K \left(c_1+t\right)}-e^{\alpha \left(c_1+t\right)}}$$
