The problem is as stated in the title, but I can't seem to get started.
My definition of BMO is $f\in\text{BMO}$ if $$\frac{1}{|Q|}\int\limits_Q |f(y)-f_Q|\, dy<\infty$$ for all cubes $Q\subset\mathbb{R}^n,$ where $f_Q$ is the average value of $f$: $$f_Q=\frac{1}{|Q|}\int_Q f(y)\, dy.$$
My definition of VMO is $f\in\text{VMO}$ if $$\frac{1}{|Q|}\int_Q|f(y)-f_Q|\, dy\rightarrow 0 $$ as $|Q|\rightarrow 0.$
I tried to use this definition of a closed set:
"A subset $F$ of $X$ is closed if, whenever $(x_n)$ is a sequence in $F$ converging to $x$, then $x$ must also be in $F$," but I can't figure out how to use this here. Any advice is helpful, thanks all.
We can equivalently work with balls, instead of cubes. Take $f\in\text{BMO}$ and a sequence $(f_n)$ in VMO so that $f_n\rightarrow f$ in BMO. Then, $$\|f_n-f\|_{\text{BMO}}=\sup_B |B|^{-1}\int_B|(f_n(y)-f(y))-(f_n-f)_B|\, dy\rightarrow 0$$ as $n\rightarrow\infty,$ where $B$ is a ball. To show that $f\in$ VMO, we must show that its average goes to zero as the radius of the balls goes to either zero or infinity. Well, \begin{align*}&|B_r|^{-1}\int_{B_r}|f(y)-f_{B_r}|\, dy\\ &\leq |B_r|^{-1}\int_{B_r}|f(y)+f_n(y)-f_n(y)+(f_n)_{B_r}-(f_n)_{B_r}-f_{B_r}|\, dy\\ &\leq |B_r|^{-1}\int_{B_r}|(f_n(y)-f(y)-(f_n-f)_{B_r}|\, dy \\ &+|B_r|^{-1}\int_{B_r}|(f_n(y)-(f_n)_{B_r}|\, dy\\ &\leq\|f-f_n\|_{\text{BMO}}+|B_r|^{-1}\int_{B_r}|(f_n(y)-(f_n)_{B_r}|\, dy. \end{align*} The first term can be made small by the BMO convergence and the second by the fact that $f_n\in\text{VMO}$ for all $n$. Thus, $f\in\text{VMO}.$
One can also define VMO as the closure of $C_0$ in BMO, in which case being closed is obvious.
FYI, the standard reference for this type of material is Stein's book on harmonic analysis. He was a whole chapter dedicated to BMO spaces.