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 $V\cap f^{-1}(U)$.
