proving that the subspace topology and the metric topology are the same My problem states:
Let $(X,d)$ be a metric space and $Y$ be a subset of $X$. Consider the function $d_Y:Y\times Y \to \mathbb{R}$, given by $d_Y(x,y)=d(x,y)$.
(a) Prove that $d_Y$ is a metric on $Y$.
(b) Prove that the topology induced by $d_Y$ on $Y$ is the subspace topology that $Y$ inherits from the metric topology on $X$.
I have finished part (a), and for part (b), I have the following:
(b) To prove that the topology induced by $d_Y$ on $Y$ ($\mathcal{T}_{d_{_Y}}$) is the subspace topology that $Y$ inherits from the metric topology on $X$ ($\mathcal{T}_Y$), we will show that $\mathcal{T}_Y = \mathcal{T}_{d_{_Y}}$.
Let $(X,d)$ be a metric space and $Y$ be a subset of $X$ with a metric $d_Y:Y\times Y \to \mathbb{R}$, given by $d_Y(x,y)=d(x,y)$, $\mathcal{T}_Y$ be the subspace topology that $Y$ inherits from $X$, $\mathcal{T}_{d_{_Y}}$ be the topology that on $Y$ induced by $d_Y$, and $\mathcal{T}_{d_{_X}}$ be the topology on $X$ induced by $d$. To prove, that $\mathcal{T}_Y = \mathcal{T}_{d_{_Y}}$, we must show that (i) the basis elements in $\mathcal{T}_Y$ are open sets in $ \mathcal{T}_{d_{_Y}}$, and that (ii) the basis elements for $\mathcal{T}_{d_{_Y}}$ are open in $\mathcal{T}_Y $.
(ii) Let $ {B}_{d_{_Y}}(x,\epsilon) \in \mathcal {B}_{d_{_Y}}$; then $ {B}_{d_{_Y}} (x, \epsilon) = \{y \in Y: d_Y(x,y) < \epsilon \}$. Since $Y \subset X$ and $x,y \in Y$, it follows that $x,y \in X$; also note that $d_Y(x,y) = d(x,y) < \epsilon$. So we have that ${B}_{d_{_Y}} (x, \epsilon)  \in \mathcal{B}_{d_{_X}}  = \{ {B}_{d_{_X}} (x, \epsilon) :x \in X, \epsilon>0\}$. But we also know that ${B}_{d_{_Y}} (x, \epsilon) \subset Y$, and therefore ${B}_{d_{_Y}} (x, \epsilon) \cap Y = {B}_{d_{_Y}} (x, \epsilon)$. Hence, as ${B}_{d_{_Y}} (x, \epsilon) \in \mathcal{B}_{d_{_X}}$ and ${B}_{d_{_Y}} (x, \epsilon) =   {B}_{d_{_Y}} (x, \epsilon) \cap Y$, it must be the case that ${B}_{d_{_Y}} (x, \epsilon)\in \mathcal{B}_Y = \{{B}_{d_{_X}} \cap Y: {B}_{d_{{_X}}} \in \mathcal{B}_{d_{_X}} \}$, which by theorem 3.5 is the basis for $\mathcal{T}_Y$. Thus, $\mathcal {B}_{d_{_Y}} \subset \mathcal {B}_Y$, which means that basis elements for $\mathcal{T}_{d_{_Y}}$ are open in $\mathcal{T}_Y $, i.e.  $\mathcal{T}_{d_{_Y}} \subset \mathcal{T}_Y $.
However, for the other direction (i), I am a bit unsure about. Unfortunately, I am not allowed to add a picture in here; I was trying to post the picture of the sets and the basis. But basically, what I have in there is the set X, inside of which I have the set Y, I have a bubble blown in $X$, $B_{d_X}(x,\epsilon)$, so that it intersects with $Y$ and $x \notin Y$. I am trying to show that basis element $B_Y = B_{d_X}(x,\epsilon) \cap Y$ is an open set in $\mathcal{T}_{d_Y}$, since it can be presented as a union of basis elements for $\mathcal{T}_{d_Y}$, i.e. union of those $B_{d_Y} (x', \delta)= \{ y' \in Y: d_Y(x',y')<\delta \}$. And since for every point $x'$ in $B_{d_X}(x,\epsilon)$ there is a $\delta>0$, such that $B_{d_X}(x',\delta) \in B_{d_X}(x,\epsilon)$, this must be possible to do, I just need to make sure that the $\delta$ bubbles are fully contained in $Y$, which is also fine, since $\delta \in \mathbb{R}^+$, and between every 2 reals there is another real; however, I am worried about the points that are in the intersection of the boundary of $Y$ and the bubble $B_{d_X}(x,\epsilon)$. How can I blow open bubbles for them that will be inside $Y$? The bubbles will have to either intersect $X$ or not quite reach $Y$'s boundary.
 A: Let $U =B_{d_X}(x;\epsilon) \cap Y \in \mathcal{T}_Y.$ Then $$U = \{y \in X \cap Y \mid d_x(x,y) < \epsilon \} = \{y \in Y \mid d_y(x,y) < \epsilon \} \in \mathcal{T}_{d_Y} $$Thus,  $\mathcal{T}_Y \subseteq \mathcal{T}_{d_Y}$. Note that reading this proof backwards gives a proof of (i),
A: If $T$ and $T'$ are topologies on a set $Y$, then to prove $T=T'$ it suffices to find a base (basis) $B$ for $T$ and a base $B'$ for $T'$ such that
(i) whenever $p\in b'\in B'$ there exists $b\in B$ with $p\in b\subset b',$ and
(ii) whenever $p\in b\in B$ there exists $b\in B'$ with $p\in b'\subset b.$
Let $(X,d)$ be a metric space and $Y\subset X.$ Let $T$ be the subspace topology on $Y$ as a subspace of $X$ and let $T'$ be the topology on $Y$ induced by the metric $d.$
Let $C$ be the set of all open $d$-balls of $X.$ Let $B=\{Y\cap c:c\in C\}.$ Now $C$ is a  base for the topology on $X$ so $B$ is a base for $T.$
Let $B'$ be the set of open $d$-balls of Y. Then $B'$ is a base for $T'$.
(i'). If $b'\in B'$ then for some  $y\in Y$ and some real $r>0$ we have  $$b'=\{z\in Y: d(y,z)<r\}=Y\cap \{z\in X:d(y,z)<r\}\in B$$ by def'n of $B.$
So $B'\subset B$. So (i) is satisfied by $b=b'.$ 
(ii'). If  $p\in b\in B $ there exists $x\in X$ and real $r>0$ such that $p\in b= Y\cap \{z\in X:d(x,z)\}=Y\cap B^X_d(x,r). $  Now $B^X_d(x,r)$ is open in $X$ so there exists real $s>0$ such that $B_d^X(p,s)\subset B_d^X(x,r).$ 
So let $b'=Y\cap B_d^X(p,s)\in B'$ and (ii) is satisfied .
Remark. In (ii') we can let $s=r-d(x,p).$ The triangle inequality then implies $B_d^X(p,s)\subset B_d^X(x,r).$ This is also used to show beforehand that  $C$ is in fact a base for a topology on $X.$
