Extending injection in Hausdorff spaces Let $f : X \to Y$ be a continous locally injective map. Then, if $f$ is injective in a compact subset, it is injective in a neighborhood of that compact. This result was proved here when  $X, Y$ are $\mathbb{R}^{n}$ . As far as I see, the result uses only metric properties, no specific property of $\mathbb{R}^{n}$is used.
Can this result be generalized to arbitrary Hausdorff spaces $X$, $Y$, by using a proof that doesn't require metric properties? I have tried to extract a cover from the locally injective neighbourhoods, but I'm not having any progress in proving global injectivity. May any of the extension theorems (Tietze like) be useful here?
Thanks in advance!
Edit: A map $f$ is locally injective in $X$ if for every point $x$ in $X$ there exists a neighborhood $U$ (a set containing an open set that contains $x$) such that $f$ is injective in $U$.
Edit2: I have started a bounty. I would be grateful if the answer (if there is any) avoids the concepts of filter and net, since those concepts are completely unfamiliar to me.
 A: The proof is somewhat complicated, but
nets and filters were avoided, as requested.
Please let me know if you have questions or found mistakes.
Let $K$ denote the compact set.
We will proceed in two steps.
Step 1:
Let $y\in f(K)$ be given.
We will show that there exists open sets $U,V$ with
$K\subset V\subset X$, $y\in U\subset Y$
such that
$$
\forall v,w\in V:
v=w \lor f(v)\neq f(w)\lor f(v)\not\in U
$$
holds.
proof:
For each point $x\in K$, we consider open neighboorhoods
$V_x$ of $x$ and $U_x$ of $y$ such that
$f$ is injective on $V_x$ and
$$
f^{-1}(U_{x}) \cap V_{x}=\emptyset
\qquad\text{if}\;f(x)\neq y.
$$
This is possible because we can separate $f(x)$ and $y$ if they are different.
Clearly, the sets $V_{x}$ cover the compact set $K$
and therefore we can extract a finite subcover, i.e.
there exists $n\in\Bbb N$, $x_1,\ldots x_n\in K$ such that
$$
K\subset \bigcup_{i=1}^n V_{x_i}.
$$
Now we define
$$
V:=\bigcup_{i=1}^n V_{x_i},
\qquad
U:=\bigcap_{i=1}^n U_{x_i}.
$$
Let us verify that $V,U$ have the desired properties.
Clearly, $V$ is an open neighboorhood of $K$ and $U$ is an open neighboorhood of $y$.
Let $v,w\in V$ be given with $f(v)\in U$ and $f(w)=f(v)$.
Then we have $v\in V_{x_i}$, $w\in V_{x_j}$ for suitable $i,j$.
Note that $v\in f^{-1}(U_{x_i})\cap V_{x_i}$ and $w\in f^{-1}(U_{x_j})\cap V_{x_j}$
hold.
From the definition of these neighboorhoods it follows that
$f(x_i)=y$ and $f(x_j)=y$ holds.
Since $f$ is injective on $K$, this implies $x_i=x_j$.
Because $f$ is also injective on $V_{x_i}=V_{x_j}$,
this implies $v=w$.
In summary, we have
$ v=w \lor f(v)\neq f(w)\lor f(v)\not\in U$
and therefore $U$,$V$ have the desired properties.
Step 2:
For each $y\in f(K)$, let $U_y\subset Y$, $V_y\subset X$ denote the neighboorhoods
$U$, $V$ from step 1.
Clearly, the sets $U_{y}$ cover the compact set $f(K)$
and therefore we can extract a finite subcover, i.e.
there exists $n\in\Bbb N$, $y_1,\ldots y_n\in f(K)$ such that
$$
f(K)\subset \bigcup_{i=1}^n U_{x_i}.
$$
Now we define
$$
V:=\bigcap_{i=1}^n V_{y_i},
\qquad
U:=\bigcup_{i=1}^n U_{y_i}.
$$
Clearly, $V$ is an open neighboorhood of $K$ and $U$ is an open neighboorhood of $f(K)$.
We claim that $f$ is injective on the open set $V\cap f^{-1}(U)$.
Let $v,w\in V\cap f^{-1}(U)$ be given with $f(v)=f(w)$.
We have $f(v)=f(w)\in U_{y_i}$ and $v,w\in V_{y_i}$ for a suitable $i$.
Recall that, by step 1, we have
$ v=w \lor f(v)\neq f(w)\lor f(v)\not\in U_{y_i} $.
This implies $v=w$, which concludes the proof that
$f$ is injective on $V\cap f^{-1}(U)$.
