That's the fluctuation-dissipation theorem. It's one of Einstein's most famous results from his "miracle year" 1905. There are two ways to answer your question. One is to derive the Fokker-Planck equation for the probability distribution function ##f(t,\vec{v})## and show that the stationary solution is the Maxwell-Boltzmann distribution with temperature ##T##, as it should be.
The other is to evaluate the expectation value of the kinetic energy and use the equipartition theorem for the long-time limit. In other words we have to calculate ##\langle \vec{v}^2(t)##. To do that we can formally solve the stochastic differential equation, using the Green's function of the "deterministic part", i.e., we look for
$$m \dot{G}(t)+\lambda G(t)=\delta(t). \qquad (*)$$
For ##t \neq 0## we have
$$G(t)=A \exp(-\gamma t), \quad \gamma=\lambda/m.$$
Since we want a "causal Green's function" we make ##G(t)=0## for ##t<0## and determine ##A## for ##t>0## such that we get the right singularity. So we integrate (*) over a small interval ##(-\epsilon,\epsilon)## and make ##\epsilon \rightarrow 0^+## to get
$$m A=1 \; \Rightarrow \; A=1/m.$$
So we have
$$G(t)=\Theta(t) \frac{1}{m} \exp(-\gamma t).$$
Then the formal solution of the Langevin equation is (for ##t>0##)
$$\vec{v}(t)=\vec{v}_0 \exp(-\gamma t) + \int_0^t \mathrm{d} t' G(t-t') \eta(t')=\vec{v}_0 \exp(-\gamma t) + \vec{v}_{\text{fluct}}(t).$$
From this we get
$$\vec{v}^2(t) = \vec{v}_0^2 \exp(-2 \gamma t) + 2 \vec{v}_0 \cdot \vec{v}_{\text{fluct}}(t) \exp(-\gamma t) + \vec{v}_{\text{fluct}}^2(t).$$
Taking the expectation value, we get because of ##\langle \vec{\eta}(t)=0##
$$\langle \vec{v}^2(t) \rangle=\vec{v}_0^2 \exp(-2 \gamma t) +\langle \vec{v}_{\text{fluct}}^2(t) \rangle.$$
Now
$$\langle \vec{v}_{\text{fluct}}^2(t) \rangle=\int_0^t \mathrm{d} t_1 \int_0^t \mathrm{d} t_2 G(t-t_1) G(t-t_2) \langle{\eta_j(t_1) \eta_j(t_2)}=\int_0^t \mathrm{d} t_1 \int_0^t \mathrm{d} t_2 6 \lambda k_{\text{B}} T \delta(t_1-t_2)G(t-t_1) G(t-t_2).$$
Evaluating the integral leads finally to
$$\langle \vec{v}_{\text{fluct}}^2(t) \rangle=\frac{3 k_{\text{B}} T}{m}[1-\exp(-2 \gamma t)].$$
For ##t \rightarrow \infty## we thus get
$$\frac{m}{2} \langle \vec{v}^2(t) \rangle \rightarrow \frac{3}{2} k_{\text{B}} T,$$
as it should be in the equilibrium limit according to the equipartition theorem. So the diffusion coefficient ##D=\lambda k_{\text{B}} T## as assumed is the correct choice. That's known as dissipation-fluctuation theorem, because it connects the friction coefficient ##\lambda## ("dissipation") to the strength of the random force in the Langevin equation, which is describing (from a macroscopic point of view) the "diffusion" of the heavy particle in the heat bath made up by the light particles.
