Your attempts are, unfortunately, flawed.
Since you know about local properties of $f$, it is better showing that $f$ is continuous at each point.
Let $x\in X$; we want to show that, for every open neighborhood $V$ of $f(x)$, there exists a neighborhood $U$ of $x$ such that $f(U)\subseteq V$. Let $U_x$ be an open neighborhood of $x$ and $V_x$ an open set in $Y$ such that $f$ induces a homeomorphism $f_{U_x}\colon U_x\to V_x$ and choose any open neighborhood $V$ of $f(x)$.
Then $V\cap V_x$ is an open set in $Y$ containing $f(x)$,
so there exists an open neighborhood $U$ of $x$ in $U_x$ such that $f(U)\subseteq V\cap V_x$; since $U$ is open in $U_x$ it is open in $X$ as well and $f(U)\subseteq V$ as requested.
Now you want to prove that $f$ is open. Let $A$ be open in $X$ and, for each $x\in A$, choose open sets $U_x\subseteq X$ and $V_x\subseteq Y$ so that $x\in U_x$ and $f$ induces a homeomorphism between $U_x$ and $V_x$.
For each $x\in A$, $f(U_x\cap A)$ is open in $V_x$, so it is open in $Y$ as well. Therefore
$$
\bigcup_{x\in A}f(U_x\cap A)
$$
equals $f(A)$ and is open in $Y$.
If $f$ is bijective, then $f^{-1}$ exists and it is continuous
because $f$ is open.
Best Answer
Here's a very detailed proof.
Let's say we have a continuous map $f:X \to Y$ of topological spaces of which we know:
In order to prove that $f$ is a homeomorphism we need to prove that $f^{-1}$ is continuous.
So, let $U' \subseteq X$ an open set and $V' = (f^{-1})^{-1}(U') = f(U')$. For each $p \in V'$ let $U_p$, $V_p$ as above (i.e. $f_{|U_p}: U_p \to V_p$ is homeomorphism), then $$ V' \cap V_p = f_{|U_p}(U' \cap U_p) $$ is open because $f_{|U_p}$ is an homeomorphism (and therefore an open map). Furthermore $$V'= \cup_{p \in V'} V' \cap V_p$$ is open, as union of open sets.