Help save proof about existence of open sets $x \in U$, $K \subset V \subset Y: \| f(u,v) - f(x,v)\| < r$, where $f: U \times V \to E$, $K$ compact 
Let $(X, \mathcal{T}_X)$ and $(Y, \mathcal{T}_Y)$ be topological spaces, $(E, \| \cdot \|)$ a normed vector space and $f: X \times Y \to E$ a continuous mapping, $K \subset Y$ a compact subset and $r > 0$ a positive number.
Show that for every $x \in X$ there exists open sets $U \subset X$ and $V \subset Y$ with $x \in U$ and $K \subset V$, so that for all $(u,v) \in U \times V$ we the strict inequality $\| f(u,v) - f(x,v) \| < r$ holds.

Note: This looks like a, if you will, "more advanced version" of this task and is made more complicated by the fact that the map $f$ is involved, so can we maybe use ideas from the proof of the previous task for solving this one?
Our proof
Let $x \in X$ and $Z := f(\{x\} \times K) \subset E$. Because $K$ is compact in $Y$ and $\{ x \}$ is compact in $X$, $\{x\} \times K$ is compact in $X \times Y$ and since $f$ is continuous, $Z$ is compact in $E$.
Now, $W := \left(W_y := B\left(y,\frac{r}{2}\right)\right)_{y \in Y}$ is an open cover of $Z$ and since $Z$ is compact, there exists and finite open subcover. Therefore, there exists a subset $Z^* \subset Z$, so that $ \bigcup_{y \in Z^*} W_y \supset Z$ is still an open cover of $Z$.
For ally $y \in Z^*$ define $U_y \subset X$ and $V_y \subset Y$ by $U_y \times V_y = f^{-1}(W_y)$.
Now here's the problem: This construction is not always possible since i.e. the open unit disk is open in $\mathbb{R}^2$ but its pre-image can't be deconstructed as a product of two sets in $\mathbb{R}$.
Our proof continues like this:
Now let $U := \cap_{y \in Z^*} U_y$ and $V := \bigcup_{y \in Z^*} V_y$
So, for all $y \in Y$ we have $x \in U_y$ and there $x \in U$, as required.
Furthermore, we have
$$
f\left( \{ x \} \times K \right)
= Z
\subset \bigcup_{y \in Z^*} W_y
= \bigcup_{y \in Z^*} f(U_y \times V_y)
= f \left( \bigcup_{y \in Z^*} U_y \times V_y \right)
= f \left( \left( \bigcup_{y \in Z^*} U_y \right) \times V \right)
$$
and so $K \subset V$.
Also, $U_y \times V_y$ is open as continuous pre-image of an open set.
Finally, $U$ and $V$ are also open as finite union or intersection of open sets.
Lastly, we show the inequality.
Let $(u,v) \in U \times V$ and $y \in Z^*$ so that $ \in V_y$.
From the way we constructed $U$ we have $(u,v) \in U_y \times V_y$ and therefore, $f(u,v) \in W_y = B\left(y, \frac{r}{2}\right)$, which is saying $\| f(u,v) - y \| < \frac{r}{2}$.
On the other hand, we have $(x,v) \in U_y \times V_y$ and so, analogously, $f(x,v) \in B\left(y, \frac{r}{2}\right)$.
By the triangle inequality we conclude
$$
\| f(u,v) - f(x,v) \|
\le \| f(u,v) - y \| + \| y - f(x,v)
< \frac{r}{2} + \frac{r}{2}
= r.
$$
Can this construction in any way be modified so everything works or is there a simpler way to approach this proof in general? Can the above in the note mentioned "easier" task be utilised in any way?
Any help is greatly appreciated.
 A: Recall that $\{U \times V : U \in \mathcal{T}_X, V \in \mathcal{T}_Y\}$ is a basis for the topology on $X \times Y$.
Since $f$ is continuous, for each $k \in K$ there exist $U_k \in \mathcal{T}_X, V_k \in \mathcal{T}_Y$ such that $(x,k) \in U_k \times V_k$ and
$$f(U_k \times V_k) \subseteq B\left(f(x,k),\frac{r}2\right)$$
$\{V_k\}_{k\in K}$ is an open cover for $K$ in $Y$ so there exist $k_1, \ldots, k_n \in K$ such that $K \subseteq \bigcup_{\ell = 1}^{n }V_{k_{\ell}}$.
Define
$$U = \bigcap_{\ell = 1}^{n} U_{k_{\ell}} \in \mathcal{T}_X
\qquad \text{and} \qquad
V = \bigcup_{\ell = 1}^{n }V_{k_{\ell}} \in \mathcal{T}_Y.
$$
Clearly, we have $x \in U$ and $K \subseteq V$.
For $(u,v) \in U \times V$ there exists $j \in \{1,\ldots,n\}$ such that $v \in V_{k_j}$. Then $(u,v) \in U_{k_j} \times V_{k_j}$ so
$$f(u,v) \in B\left((x,k_j),\frac{r}2\right)$$
Also $(x,v) \in U_{k_j} \times V_{k_j}$ so $f(x,v) \in B\left((x,k_j),\frac{r}2\right)$ as well.
Then
$$\|f(u,v) - f(x,v)\| \le \|f(u,v) - f(x,k_{j})\| + \|f(x,k_j) - f(x,v)\| < \frac{r}2 + \frac{r}2= r$$
