Prove that the subspace topology is the coarsest topology on $Y$ for which the inclusion function $i:Y \longrightarrow X$ given by $i(y)=y$ is continuous.
Attempt
Define $T^{\prime}=\lbrace V\subset X \mid i^{-1}(V) \text{ is open in $Y$} \rbrace$
since $Y$ have the subspace topology $ i^{-1}(V)$ should look like $Y \cap U$ where $U$ is open in $X$.
By definition $i^{-1}(V)=\lbrace y\in
Y \mid i(y)\in V\rbrace=\lbrace y\in Y \mid y\in V\rbrace$ since $V\subset X$ and not necessary in all $Y$ we conclude that $T^{\prime}=\lbrace Y\cap V \mid V\subset X \rbrace$ and since the subspace topology in $Y$ is given by $T=\lbrace U\cap Y \mid U\, \text{is open in $X$} \rbrace$
And obviously $T\subset T^{\prime}$ and therefore the subspace topology on $Y$ is coarsest topology for which the function $i$ is continuous.
Best Answer
Your attempt contradicts itself: you have two descriptions of $T'$ that clearly cannot be equal. On the one hand you let
$$T'=\{V\subseteq X:i^{-1}[V]\text{ is open in }Y\}\,;\tag{1}$$
clearly $X\in T'$, since $i^{-1}[X]=Y$. But later you conclude that
$$T'=\{Y\cap V:V\subseteq X\}\,,\tag{2}$$
which contains only subsets of $Y$; if $Y\subsetneqq X$, then $X$ is not in this version of $T'$, so $(1)$ and $(2)$ cannot describe the same family of sets. And there is a further problem with $(2)$: it is simply $\wp(Y)$, since every subset of $Y$ has the form $Y\cap V$ for some $V\subseteq X$.
Let $\tau_X$ be the topology on $X$, $\tau_Y$ the subspace topology on $Y$, and $\tau$ the coarsest topology on $Y$ for which $i$ is continuous; by definition
$$\tau=\left\{i^{-1}[V]:V\in\tau_X\right\}\,,$$
and you want to show that $\tau=\tau_Y$. You know that $\tau_Y=\{V\cap Y:V\in\tau_X\}$, so you want to show that
$$\left\{i^{-1}[V]:V\in\tau_X\right\}=\{V\cap Y:V\in\tau_X\}\,.$$
The natural way to try to do this is to try to prove that $i^{-1}[V]=V\cap Y$ for each $V\in\tau_X$. This is true and is not hard to prove; see if you can manage it.