Closure of a subset of a subspace of a topological space

Hint: Use the characterization of the closure of $T$ in $X$ as $$\overline{T}=\bigcap_{\substack{\text{closed}\, C\,\subseteq X\\ C\,\supseteq T}} C$$ together with the definition of the subspace topology on $Y$ to prove that the statement is true, i.e. that the closure of $T$ in $Y$ is $\overline{T}\cap Y$.

Suppose $x$ is in $\overline{T}^{(Y)}$, the closure of $T$ in $Y$. This means in particular that $x \in Y$, obviously. If $O$ is an open subset of $X$ that contains $x$, then $O \cap Y$ is an open subset of $Y$ that contains $x$ and so $(O \cap Y) \cap T \neq \emptyset$, so in particular $O \cap T \neq \emptyset$. Hence every open neighbourhood (in $X$) of $x$ intersects $T$, so $x \in \overline{T}$. So $\overline{T}^{(Y)} \subset Y \cap \overline{T}$.

The other implication is also not too hard: suppose $x$ is in $Y \cap \overline{T}$. Let $O$ be any open neighbourhood of $x$ in $Y$. By the definition of the subspace topology this means that $O = U \cap Y$ for some open subset $U$ of $X$. As $x \in \overline{T}$, $U$ intersects $T$, so (recall that $T \subset Y$): $\emptyset \neq U \cap T = U \cap (T \cap Y) = (U \cap Y) \cap T = O \cap T$. So we have shown that every open neighbourhood (in $Y$) of $x$ intersects $T$, so $Y \cap \overline{T} \subset \overline{T}^{(Y)}$.

So we have the required equality.