If $Y \subset R^n$ has the subspace topology from the standard topology, then $U \subset Y$ iff If $Y \subset R^n$ has the subspace topology from the standard topology, then $U \subseteq Y$ is an open subset within Y if and only if given any $x \in U$ there exists $\delta >0 $ such that $$(B_{\delta}(x) \cap Y )\subseteq U$$
Left side $\implies$ right side:
Assume $U \subseteq Y$ is open within Y. Then there exists an open $V within R^n$ such that $V \cap Y = U$ (follows from the definition of a subspace topology on the standard topology on $R^n$).
And by following the same convention : 
$$(U \cap Y )\subseteq U$$ since $U \subseteq R^n$
Since $U$ is an open set, given any $x \in U$ $\exists \delta >0$ such that $B_{\delta}(x) \subseteq U$ Then
$$(B_{\delta}(x) \cap Y) \subseteq U$$.
Do you think my proof makes sense? I'm not so sure about the procedure I followed so please leave your comments below. Also, I couldn't prove the converse. Any help/hint will be appreciated. 
Thanks
 A: You can't say that $U$ is an open set in the space.
Take for instance $\mathbb{R}$ with the usual topology and $Y=[0,5]$
Then the set $(2,5]=Y \cap (2,6)$ where $(2,6)$ is open in $\mathbb{R}$
but $(2,5]$ is not open in $\mathbb{R}$ with respect to the usual topology
Now to prove the statement:
Because $U$ is open in $Y$ exists $V$ open in $\mathbb{R}^n$ such that $U=Y \cap V$
Now if $x \in U$ then $x \in V$ thus exists $\delta>0$ such that $B_{\delta}(x) \subseteq V$ because $V$ is open in the space
thus $ B_{\delta}(x) \cap Y   \subseteq  V \cap Y =U$
A: Proof: $(\implies)$ Suppose $U \subseteq Y$ is open in $Y$, then by definition $U = V \cap Y$ for some open set $V$ in $\mathbb{R}^n$. Choose $x \in U$ to be arbitrary, then $x \in V$ also. Since $V$ is open in $\mathbb{R}^n$, there exists an open ball $B_{(\mathbb{R}^n, d)}(x, \delta) \subseteq V$, therefore by taking intersections with $Y$, we can see there exists a $\delta > 0$ such that $B_{(\mathbb{R}^n, d)}(x, \delta) \cap Y \subseteq V \cap Y = U$.
Conversely $(\impliedby)$ suppose there exists a $\delta > 0$ such that for any $x \in U$ we have $B_{(\mathbb{R}^n, d)}(x, \delta) \cap Y \subseteq Y$, we show that $U$ is open in $Y$.
Now recall that in a general topological space $X$ if $A \subseteq X$ and if for each $x \in A$, there is an open set $W$ in $X$ containing $x$ such that $W \subseteq A$, then $A$ is open in $X$. 
In the above, let $U =A$, and let $Y = X$, with $W = B_{(\mathbb{R}^n, d)}(x, \delta) \cap Y$ being our open set in $Y$, and since $W \subseteq Y$, and $x \in U$ was arbitrary we can conclude $U$ is open in $Y$. $\square$
