In order to show that it is bounded I need to show that it is contained in some neighbourhood of $\mathbb{R}$. Here is my proof:

Let $\mathcal{A} = \{ A\_1,\ldots,A\_n \}$ a finite collection of bounded sets in $\mathbb{R}$, since each $A\_i$ ($i=1,...,n$) is bounded and because of the greatest lower bound axiom, it has a least upper bound and a greatest lower bound, let $p=\max(\sup A\_1,\ldots,\sup A\_n)$ and $q=\min(\inf A\_1,\ldots,\inf A\_n)$, define $y = \frac{p+q}{2}, \ \alpha = \frac{p-q}{2} $. Let $x \in \bigcup\_{j=1}^n A\_j$, then $x \in A\_k$ for at least one $k$ (where $1\leqslant k \leqslant n$). Let $\epsilon > 0$, consider the neighbourhood $B(y,\alpha) = (q-\epsilon, p+ \epsilon)$, then: $q-\epsilon < \inf A\_k \leqslant x \leqslant \sup A\_k < p + \epsilon$, so that $x \in B(y, \alpha)$, then $A\_k \subset B(y, \alpha), \ \forall k=1,\ldots,n, \ \forall \epsilon >0$ (since $x$ was arbitrary). Therefore the finite union of bounded sets in $\mathbb{R}$ is bounded.

I think this successfully proves the theorem, but I am not 100% sure, can some one check my proof? Thanks!!