Show that every local homeomorphism is continuous and open therefore bijective local homeomorphism is a homeomorphism

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.

Allow me to add another answer for proving the continuity. The difference from @egreg answer is only in looking at proof from a different angle.

Let $U \subseteq Y$ be open in $Y$. We must show that $f^{-1}(U)$ is open in $X$. Let $x \in f^{-1}(U)$ be arbitrary.

By definition of local homeomorphism, $\exists\ V_x \subseteq X$ which is a neighbourhood of $x$ such that $f(V_x)$ is open in $Y$ and $f\big\vert_{V_x}:V_x\rightarrow f(V_x)$ is a homeomorphism.

Since $U$ and $f(V_x)$ are open in $Y$, then, so is their intersection $U \cap f(V_x)$ is open in $Y$.

Also, continuity of $f\big\vert_{V_x}$ implies that,

$$f\big\vert_{V_x}^{-1}(U \cap f(V_x)) = \{x \in V_x: f(x) \in U \cap f(V_x)\} = V_x \cap f^{-1}(U)$$

is open in $X$. But $V_x \cap f^{-1}(U)$ is a neighbourhood of $x$ contained in $f^{-1}(U)$. Because $x$ is an arbitrary point in $f^{-1}(U)$, therefore,

$$f^{-1}(U) = \bigcup\limits_{x \in f^{-1}(U)}(V_x\cap f^{-1}(U))$$

is an arbitrary union of open subsets of $X$, hence, is open in $X$. Therefore, $f$ is continuous.