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?

 A: 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.
A: 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 $Y$.
A: 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.

