Analysis Problem: Prove $f$ is bounded on $I$ Let $I=[a,b]$ and let $f:I\to {\mathbb R}$ be a (not necessarily continuous) function with the property that for every $x∈I$, the function $f$ is bounded on a neighborhood $V_{d_x}(x)$ of $x$. Prove that $f$ is bounded on $I$.
Thus far I have that,
For all $n∈I$ there exist $x_n∈[a,b]$ such that $|f(x_n)|>n$. By the Bolzano Weierstrass theorem since $I$ is bounded we have the sequence $X=(x_n)$ is bounded. This implies there is a convergent sub-sequence $X'=(x_{n_r})$ of $X$ that converges to $c$, $c∈[a,b]$. Since $I$ is closed and the element of $X'$ belongs to $I$, it follows from a previous theorem that I proved that $c∈I$. Here is where I get stuck, I want to use that the function $f$ is bounded on a neighborhood $V_{d_x}(x)$ somehow to show that $f$ is bounded on $I$. I'm not sure how to proceed.
$f$ is bounded on $I$ means if there exist a d-neighborhood $V_d(c)$ of $c$ and a constant $M>0$ such that we have $|f(x)|\leq M$ for all $x$ in $A ∩  V_d(c)$.
I would like to do try a proof by contradiction somehow.

So, by hypothesis, you have an open ball $V_d(c)$ and $M$ such that $|f(x)|\le M$ for all $x\in I\cap V_d(c)$ (I think you meant $I$ here, not $V$).  But you know that the sequence $(x_{n_r})$ converges to $c$. This means that there is an $m$ so that $x_{n_r}\in I\cap V_d(c)$ whenever $r>m$. Now from the way you selected the $x_n$, what can you conclude?
