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

Follow up on another question I asked recently: Topology: Show restriction of continuous function is continuous, and restriction of a homeomorphism is a homeomorphism

Definition: Let $(X, \mathcal{T})$ and $(Y, \mathcal{J})$ be topological spaces. A function ${\displaystyle f:X\to Y\,}$ is a local homeomorphism if for every point $x \in X$ there exists an open set $U \subseteq X$ containing $x$ and an open set $V \subseteq Y$ such that the restriction ${\displaystyle f|_{U}:U\to V\,}$ is a homeomorphism.

This definition is a bit alarming because it starts with "...if for every point $x \in X$ there exists an open set $U \in \mathcal{T}$...", makes it seem like a property of the underlying space. Can we always find an open $U$? But anyways.

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

Proof: (Honestly not sure what I am doing but proceed regardless)

Let $(X, \mathcal{T})$ and $(Y, \mathcal{J})$ be topological spaces and function ${\displaystyle f:X\to Y\,}$ is a local homeomorphism. We will show that $f$ is continuous and open.

First show $f$ is continuous.

$f$ is continuous if for all $V \in \mathcal{J}, f^{-1}(V) \in \mathcal{T}$. Take some $V \in \mathcal{J}$, then $V$ is a subspace equipped with subspace topology $\mathcal{J}_V = \{V \cap W| W \in \mathcal{J}\}$.

Consider the inverse of the restriction $f^{-1}|_U$ on an open set in $\mathcal{J}_V$, then $f^{-1}|_U(V \cap W) = f^{-1}|_U(V) \cap f^{-1}|_U(W) $$= f^{-1}(V) \cap U \cap f^{-1}(W) \cap U = f^{-1}(V) \cap f^{-1}(W) \cap U$.

Then $f^{-1}(V) = f^{-1}(W) \cup U \cup f^{-1}|_U(V \cap W)$. We note all the sets on the right hand side are open. In particular, $U$ is open, $f^{-1}|_U(V \cap W)$ is open by definition of homeomorphism (?? $f^{-1}(W)$ ??), hence $f$ is continuous. ($\leftarrow$ something wrong here!)

Next show $f$ is open.

$f$ is open if $\forall U \in \mathcal{T}, f(U) \in \mathcal{J}$. Consider the restriction $f|_U$ on the subspace topology on $U$, $\mathcal{T}_U = \{U \cap M | M \in \mathcal{T}\}$. $f|_U(U \cap M) = f|_U(U) \cap f|_U(M) = V \cap f(M) \cap f(U)$

Then $f$ is open since $f(U) = f|_U(U \cap M) \cup V \cup f(M)$ and $f|_U(U \cap M)$ is open by definition of homeomorphism, $V$ is open in $\mathcal{T}$ (?? $\cup f(M)$ ??) ($\Leftarrow$ another mistake here)

I'm not quite sure how to proceed with showing bijective + continuous + open + local = homeomorphism part.

Can someone help me fix those two problems and give me some ideas how to conclude that bijective local homeomorphisms are homeomorphisms?

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.

Claim 1: Every local homeomorphism is an open map.

Proof: Let $f:X\rightarrow Y$ be a local homeomorphism. Let $U\subseteq X$ be open. If $x\in U$ then there exists some open subset $V\subseteq X$ such that on $V$, $f$ is open onto an open subset of $Y$. Hence $f(U\cap V)$ is open in $Y$. Note, $f(x)\in f(U\cap V)\subseteq f(U)$. Thus we may conclude that $f(U)$ is the union of open sets and is therefore open.

Claim 2: Every bijective local homeomorphism is a homeomorphism.

Proof: Let $f:X\rightarrow Y$ be a bijective local homeomorphism. Let $V$ be open in $Y$. We must show that $f^{-1}(V)$ is open in $X$. Let $x\in f^{-1}(V)$. Let $U_{x}$ be an open neighborhood of $x$ such that $f|_{U_x}: U_x\rightarrow f(U_x)$ is a homeomorphism onto an open subset of $Y$. Hence, $f(x)\in f(U_x)\cap V$. Choose an open set $W_{f(x)} \owns f(x)$ such that $W_{f(x)}\subseteq f(U_x)\cap V$. Then, $f^{-1}(W_{f(x)})$ is open in $X$. Moreover, $f^{-1}(W_{f(x)})\subseteq f^{-1}(V)$ . Thus $f^{-1}(V)$ is open. Now note, if $f:X\rightarrow Y$ is a bijective local homeomorphism, then $f^{-1}: Y\rightarrow X$ is a bijective local homeomorphism. Hence repeating a similar argument as we did for $f$ allows us to conclude that $f$ is a homeomorphism. Alternatively, if $U$ is open in $X$ then $f(U)$ is open in $Y$. Hence, $(f^{-1})^{-1}(U)=f(U)$ is open in $Y$.