compactness requirement for the tube lemma of a product space. The tube lemma: Let $X,Y$ be topological spaces s.t $Y$ is compact. Let $X_0 \in X$ and let $N$ be an open set in $X \times Y$ so that $x_0 \times Y$ is contained in $N$. Then there exits a neighborhood $W$ of $x_0$ such that $W \times Y \subseteq N$.
I understand the proof for this lemma, and the fact that it relies on $W=U_1 \cap...\cap U_2$ for finitely many basis elements of $x_0 \times Y$ (which is compact by homeomorphism.)
So, I can see formally how this proof relies on the finite intersection of basis elements of a compact $Y$.
However, my intuition does not follow from this. Is it the fact that the intersection of infinitely many $U_i$ need not be open? Or that the infinite intersection maybe trivial?
Munkres provides a counter-example, but I simply do not  understand it.
 A: Let $X = (-1,1)$ and $Y = (1, \infty)$ and $x_0 = 0$. Consider the set
$$ N = \{ (x,y) \, | \, 0 < |x| < 1, 1 < y < \frac{1}{|x|} \} \cup (\{x_0\} \times Y). $$
Drawing $N$, we see that $N$ looks like

The set $N$ is an open set containing (part of) the $y$-axis $\{ 0 \} \times (1, \infty) = \{ x_0 \} \times Y$ and it becomes more narrow as $y$ approaches infinity. For each $0 < \varepsilon < 1$, the open rectangle $(-\varepsilon, \varepsilon) \times (1, \frac{1}{\varepsilon})$ is contained in $N$ so given a point $(0, y_0)$ with $y_0 > 1$, we can choose $\varepsilon$ with $0 < \varepsilon < \frac{1}{y_0}$ and obtain an open product neighborhood $U_{y_0}$ around $(0,y_0)$ that is contained in $N$. However, as $y_0 \to \infty$, the base of such open product neighborhood must become smaller and smaller and the set $N$ does not contain an open set of the form $(-\varepsilon, \varepsilon) \times Y$. The $U_{y_0} \cap Y$ form an open cover of $Y$ but they do not have a finite subcover from which one could try and extract a neighborhood of the form $(-\varepsilon, \varepsilon) \times (1, \infty)$.
