I was solving exercises on real analysis by Introduction to Real Analysis by Robert Bartle and Donald Sherber forth edition exercise 5.3 page 141 and I am confused what does $V_{\delta_x}(x)$ in
Let $I :=[a,b]$ and let $f : I \to \mathbb{R}$ be a (not necessarily continuous) function with the property that for every $x \in I$, the function $f$ is bounded on a neighborhood $V_{\delta_x}(x)$of x (in the sense of Definition 4.2.1). Prove that $f$ is bounded on $I$
To add more context
4.2.1 Definition: Let $A \subset \mathbb{R}$, let $f : A \to \mathbb{R}$, and let $c \in \mathbb{R}$ be a cluster point of $A$. We say that $f$ is bounded on a neighborhood of $c$ if there exists a $\delta$-neighborhood $V_{\delta}(c)$ of $c$ and a constant $M > 0$ such that we have $|f(x)|\le M$ for all $x \in A \cap V_{\delta}(c)$.
I know the notation of neighborhood $V_{\delta}(x)$ but what exactly does the extra $x$ denote
The $x$ in the $\delta_x$ implies a dependence of the $x$ you are using. It may be that you'll need different deltas. Some autors also write a dependence of $\epsilon$ when they are proving limits ( $\forall \epsilon >0 \exists \delta_{\epsilon} ...$ ) and other definitions.
As for the proof, you can demonstrate it using the Heine-Borel property of sets in $\mathbb{R}$. Notice that $\lbrace V_{\delta_{x}} (x) | x \in \left[ a,b \right]\rbrace$ is an open cover of $\left[ a,b \right]$. Since $ \left[ a,b \right]$ is compact in $\mathbb{R}$, by the Heine-Borel property, there $\exists$ an open finite subcover of $\left[ a,b \right]$, say, $\lbrace V_{\delta_{x_{i}}} (x_i) | x_{i} \in \left[ a,b \right] (1 \le i \le n) \rbrace$.
By the hypothesis property, $f$ is bounded on every neighbourghood of the mentioned subcover. So, $\forall i \in \lbrace 1,..., n \rbrace$, $\exists M_i > 0 $ such that $ |f(x)| \leq M_i $ $\forall x \in V_{\delta_{x_{i}}} (x_i) \cap A $.
Choose $M=max \lbrace M_i | 1 \leq i \leq n\rbrace >0$. This $M$ exists and its well defined since $\lbrace M_1 , ... M_n \rbrace$ is a finite set. Then, $\forall x \in A = \bigcup_{i=1}^{n} \left( A \cap V_{\delta_{x_{i}}} (x_i) \right)$ $|f(x)| \leq M$. Therefore $f$ is bounded on $ \left[ a,b \right]$.