> For any given $T>0$, $y(t,\epsilon)\rightarrow x(t)$ in probability as
> $\epsilon\rightarrow0$ uniformly for $t\in[0,T]$


Proof: I will sketch the main steps, then fill in the details later.

$(\text{Eq}.(3)-\text{Eq}.(1))/\epsilon$ gives
$$dy=-(k_0y+k_1(x_0-1)+\epsilon k_1y)\,dt+(\eta_0y+\eta_1x_0+\epsilon \eta_1y)\,dB \tag5$$
$\text{Eq}.(5)-\text{Eq}.(4)$ gives
$$dz = -(k_0z+\epsilon k_1y)\,dt+(\eta_0z+\epsilon \eta_1y)\,dB \tag6$$
where $z=y-x_1$.

I am going to prove for any given $T>0$,
$$
\mathbf E\big[y(t,\epsilon)^2\big]<M\quad\text{and}\quad\mathbf E\big[z(t,\epsilon)^2\big]<\epsilon^2Be^{AT},\quad \forall\epsilon\in(0,\epsilon_0),\ t\in[0,T] \tag7
$$
for some positive constants $\epsilon_0,\,M,\,A,\,B$. Therefore by the Chebyshev-Markov inequality, we have $z(t,\omega,\epsilon)\rightarrow0$ in probability as $\epsilon\rightarrow0$ uniformly for $t\in[0,T]$.

We will show the derivation for the inequality for $z$ in Eq.(7). That for $y$ is similar. 

Take the integral form of Eq. (6), square it and apply the Cauchy-Schwartz inequality.
$$\frac14z(t)^2\le\Big(\int_0^t k_0z\,ds\Big)^2+\Big(\int_0^t\epsilon k_1y\,ds\Big)^2+\Big(\int_0^t\eta_0z\,dB_s\Big)^2+\Big(\int_0^t\epsilon \eta_1y\,dB_s\Big)^2 \tag8
$$
For a deterministic function $a(t)$ and a stochastic funciton $u(t,\omega)$ where $\omega$ is an element of the sample space,
$$\mathbf E\Big[\Big(\int_0^t a(s)u(s,\omega)dB(s,\omega)\Big)^2\Big]=\mathbf E\Big[\int_0^t (a(s)u(s,\omega))^2\,ds\Big]\le \int_0^t a(s)^2ds\, \int_0^t \mathbf E[u(s,\omega)^2]ds \tag9$$
The last inequality is from Cauchy-Schwartz.

Take expectation on Eq.(8) and apply Eq.(9) and again Cauchy-Schwartz, we have
$$\mathbf E[z(t)^2]\le \alpha(t)\int_0^t\mathbf E[z(s)^2]\,ds +\epsilon^2\beta(t)\int_0^t\mathbf E[y(s)^2]\,ds$$
for some positive nondecreasing continuous deterministic funtions $\alpha(t)$ and $\beta(t)$. Let $v(t):=\int_0^t\mathbf E[z(s)^2]\,ds$, we have
$$v$$