The way I understand it, homogeneity of Space means that the Lagrangian possess translational Invariance i.e. doesn't change when you do a transformation of the type:
\begin{equation}
\tilde{x}^{i}=x^{i}+\epsilon^{i}
\end{equation}
Then from Noether's Theorem, you can derive the following:
\begin{equation}
\begin{aligned}
0 &=\delta S \\
&=\sum_{i} \int_{t_{1}}^{t_{2}}\left(\frac{\partial L}{\partial x^{i}} \delta x^{i}+\frac{\partial L}{\partial \dot{x}^{i}} \delta \dot{x}^{i}\right) d t \\
&=\sum_{i} \int_{t_{1}}^{t_{2}}\left(\frac{\partial L}{\partial x^{i}} \delta x^{i}+\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{x}^{i}} \delta x^{i}\right)-\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{x}^{i}}\right) \delta x^{i}\right) d t \\
&=\left.\sum_{i} \frac{\partial L}{\partial \dot{x}^{i}} \varepsilon^{i}\right|_{t_{1}} ^{t_{2}}+\sum_{i} \int_{t_{1}}^{t_{2}}\left(\frac{\partial L}{\partial x^{i}}-\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{x}^{i}}\right)\right) \varepsilon^{i} d t
\end{aligned}
\end{equation}
Taken that your particle obeys the Euler-Lagrange equation the integral vanishes. Since $t_1,t_2$ and $\epsilon^i$ are all arbitrary we can conclude that:
\begin{equation}
\frac{\partial L}{\partial \dot{x}^i}=0
\end{equation}
but that's just the momentum of the particle, therefore momentum is conserved.
I'm gonna try to also give an intuitive explanation. Let's say you throw a stone off a cliff. Gravity makes each height unique and the motion of the stone will depend on the height of the cliff you chose. If you try to do the same experiment on a nearby cliff with the same height you will get the exact same result as the previous one. Now think of the stone, just like you wouldn't be able to distinguish one cliff from another, the stone can't distinguish nearby horizontal points but it can distinguish nearby vertical points. We know from Newton's equation that if you do the experiment, momentum changes in the vertical direction, the case where nearby points are distinguishable for the stone. Therefore we expect that in the horizontal direction, where nearby points are indistinguishable, momentum is conserved.