Always existence of a smaller neighborhood. An Idea that is always used by my professor that I do not know how to prove it:
If $X$ is topological space and $x\in X$ and $\{x\} \times I \subset V$ where $V$ is open in $X \times I,$ why we are always sure that there exists a nhbd $U$ of $x$ in $X$ such that $\{x\} \times I \subset U \times I \subset V.$
Is there a prove for this fact, specifically, we do not have any extra condition on our topological space.
My idea is:
1-Since $V$ is open, then every point in $V$ (including $x$) has a nhbd, say $B(r; x)$ lying in $V.$ We are sure that there exists $B(r/2; x)$ lying in a smaller neighborhood of $V$ call it $U.$
Is this a proof or I should add more stuff? or is it wrong?
2- What if we replace $I$ the unit interval by the real line $\mathbb R$?
 A: As Anne Bauval suggests in the comments, you should look up the tube lemma, which is the desired statement with the unit interval $I$ replaced more generally with any compact space $Y$.
I want to address your second question, which provides some intuition for why we want compactness for the space $Y$ we are producting with.
Let $X = \mathbb{R}$, and consider $V \subset \mathbb{R} \times \mathbb{R}$ defined by
$$V = \{(x, y)  \mid -e^{y} < x < e^{y}\}.$$ Check for yourself that this is an open set.
Consider $\{0\} \times \mathbb{R} \subset V.$ We claim there doesn't exist an open set $U \subset \mathbb{R}$ containing $\{0\}$ such that $$\{0\} \times \mathbb{R} \subset U \times I \subset V.$$ To see this, note that any open set $U$ of $\mathbb{R}$ containing $\{0\}$ must contain $(-\varepsilon, \varepsilon)$ for some $\varepsilon > 0$. This means that $V$ must contain $(-\varepsilon, \varepsilon) \times \mathbb{R}$. But for small enough $y$ we have $e^{y} < \varepsilon$, so this is impossible!
So the claim doesn't hold when $I$ is replaced with $\mathbb{R}$.
