How to prove this statement for metric subspaces? How to prove the following statement about an intersection of two open sets in a metric subspace?

If $(Y,d) \subset (X,d)$ are metric space and its subspace relatively, then $$\forall G_1, G_2 \subseteq (Y,d) : G_1 \cap G_2 = \varnothing$$
$$\exists H_1, H_2 \subseteq (X,d) : H_1 \supseteq G_1, H_2 \supseteq G_2, H_1 \cap H_2 = \varnothing$$

It is supposed that metric spaces topic in the course I read are introduced before any other topological terms, so I assume that proof must be simple. 
Thank you in advance!
 A: Following the proof given in Metric spaces are completely normal, we first associate to each $p \in G_1$ a positive number $\delta_p$ such that the open ball in $Y$
$$B_Y(p,\delta_p) := \{y \in Y : d(p,y) < \delta_p\} \subseteq G_1$$
Similarly, to each $p \in G_2,$ we associate a positive number $\delta_p$ such that $B_Y(p,\delta_p) \subseteq G_2$ (since $G_1 \cap G_2 = \emptyset,$ there's no problem using the same notation for both)
Then we define $$H_1 = \bigcup_{p\in G_1} B_X\left(p,\frac{\delta_p}{2}\right) \\ H_2 = \bigcup_{p\in G_2} B_X\left(p,\frac{\delta_p}{2}\right)$$
By definition, these are open sets in $X$ satisfying $G_i \subseteq H_i$ for $i=1,2.$  Therefore it suffices to show that $H_1\cap H_2 = \emptyset.$
Towards a contradiction, suppose they aren't disjoint and choose some $x \in H_1 \cap H_2.$  Then there are $g_1 \in G_1, g_2 \in G_2$ such that $$x \in B_X\left(g_1,\frac{\delta_{g_1}}{2}\right) \cap B_X\left(g_2,\frac{\delta_{g_2}}{2}\right) \\d(g_1,g_2) \leq d(g_1,x) + d(g_2,x) < \frac{\delta_{g_1}}{2} + \frac{\delta_{g_2}}{2} \leq \max\{\delta_{g_1}, \delta_{g_2}\}$$
This then shows that $G_1 \cap G_2 \neq \emptyset,$ which is our contradiction, so we may conclude $H_1 \cap H_2 = \emptyset,$ finishing the proof.
