$f: X \rightarrow Y$ injective, closed. Prove that for $U \in \mathcal{T}_x$ and $M$ ⊆ $Y$, ∃ $V \in \mathcal{T}_y$ s.t $M$⊆$V$ and $f^{-1}(V)$ ⊆ $U$

Question

Let $(X,\mathcal{T}_x), (Y,\mathcal{T}_y)$ be topological spaces and $f: X \rightarrow Y$ be an injective and closed map (i.e. it maps closed sets to closed sets). Let $U$ be an open subset of $X$ and $M$ be a non-empty subset of $Y$ such that $f^{−1}(M) \subset U$. Prove that there exists an open subset $V$ of $Y$ such that $M \subset V$ and $f^{−1}(V) \subset U$

I've made three attempts below (or rather two half-attempts and one real attempt). I've placed exclamation marks on top of equalities or bolded parts where I suspect my logic might be incorrect.

1. My initial idea was that since $f$ is a closed map, it means that $f^{-1}: (Y,\mathcal{T}_y) \rightarrow (X,\mathcal{T}_x)$ is continuous since: $ (f^{-1})^{-1}(W) = f(W)$ is closed in $\mathcal{T}_y$ for every $W$ closed in $\mathcal{T}_x$. But this would then imply that $f(U)$ is open in $\mathcal{T}_y$ for every $U$ open in $\mathcal{T}_x$ (since $f^{-1}$ is continuous), and this cannot be true, otherwise an open map would be equivalent to a closed map, which is not true from what I know.

2. If it somehow instead turned out that $f$ is continuous instead, then maybe one could do the proof by: $U \subseteq \mathcal{T}_x \iff \exists V \subseteq \mathcal{T}_y$ s.t $f^{-1}(V) = U$. Since $f^{-1}(M) \subseteq U$ and $f^{-1}(V) = U \Rightarrow M \subseteq V, f^{-1}(V) \subseteq U$ and $V$ open.

3. $U$ open in $\mathcal{T}_x \Rightarrow X \setminus U$ closed in $\mathcal{T}_x \Rightarrow f(X \setminus U)$ closed in $\mathcal{T}_y$. But $f(X \setminus U) = f(X) \setminus f(U) \overset{!}{=} Y \setminus f(U)$ open in $\mathcal{T}_y \Rightarrow f(U)$ open in $\mathcal{T}_y$. $f(U)$ is our desired $V$ since: $f^{-1}(V) = f^{-1}(f(U)) = U$ (hence $f^{-1}(f(U)) \subseteq U$), $M \subseteq V = f(U)$

My questions are:

1. In attempt 1, I've assumed that $f^{-1}(W)$ exists for every $W \in \mathcal{T}_y$, which can only be the case for surjective $f$. Is this where my contradiction is arising from?
2. Am I on the right track in my second attempt, if so how do I show $f$ is continuous? However, it seems a bit odd that I'm getting $f^{-1}(V) = U$ instead of $f^{-1}(V) \subseteq U$.
3. What about attempt 3? Again here I've assumed that $f$ is surjective since I set $f(X) = Y$.
4. If none of these attempts are on the right path, could you give me a hint?