Suppose that $f$ is not bounded on $I$. Then for each $n\in\Bbb N$ there is an $x_n\in I$ such that $|f(x_n)|\ge n$. The sequence $\langle x_n:n\in\Bbb N\rangle$ is bounded, since it lies in $I$, so it has a convergent subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$. Let $x$ be the limit of this subsequence. There is an $m\in\Bbb N$ such that $x_{n_k}\in V_{\delta_x}$ for every $k\ge m$, contradicting the boundedness of $f$ on $V_{\delta_x}$.
Added: Recall that the points $x_n$ were chosen so that $|f(x_n)|\ge n$ for each $n\in\Bbb N$. In particular, $|f(x_{n_k})|\ge n_k$ for all $k\in\Bbb N$. Now let $M$ be any positive real number. There is a $k_0\in\Bbb N$ such that $n_{k_0}\ge M$. (Recall that $\langle x_{n_k}:k\in\Bbb N\rangle$ is a subsequence of $\langle x_n:n\in\Bbb N\rangle$, so $\langle n_k:k\in\Bbb N\rangle$ must be strictly increasing.) Now choose any $k\ge\max\{m,k_0\}$; then $n_k\ge n_{k_0}\ge M$, and $x_{n_k}\in V_{\delta_x}$ (because $k\ge m$), so $x_{n_k}$ is a point of $V_{\delta_x}$, and $|f(x_{n_k})|\ge M$. $M$ was arbitrary, so there are points of $V_{\delta_x}$ at which $f$ assumes values with arbitrarily large magnitudes $-$ which is exactly what it means for $f$ to be unbounded on $V_{\delta_x}$.