Let $X$ be a topological space, and $I=[0,1]$. Consider $X \times I$ with the product topology. Now fix $x_0\in X$ and $U\subset X\times I$ an open that contains $\{x_0\}\times I$. Prove that there exists an open $V\subset X$ so that $\{x_0\}\times I\subset V\times I\subset U$
My attempt:
Since U is an open set in the product topology, $U=\bigcup( U_{x_0}\times I_i)$ where $I_i$ are open sets of I and $U_{x_0}$ is an open set of $X$ that contains $x_0$. Since I is compact $W$ is the finite union $W=\bigcup( U_{x_0}\times I_i)\subset U$.
Now I would need to find a subset of $U_{x_0}$ that also contained $x_0$, how?
We suppose here that the topology used on $I$ is the stadard one
As $U$ is an open that contains $\{x_0\}\times I$, for all $i \in I$, there is an open $U_i \subseteq X$ that contains $x_0$ and an open $J_i \subseteq [0,1]$ that contains $i$ such that $(x_0,i) \in U_i \times J_i$.
$\mathcal J = \{J_i \mid i \in I\}$ is an open cover of $I$. Using the introduction remark, $I$ is compact and we can extract from $\mathcal J$ a finite subcover $\{J_{i_1}, \dots ,J_{i_n}\}$.
$V=\cap_{j=1}^n U_{i_j}$ is an open subset of $X$ that satisfies the conditions we're looking for.