Vtyjkyuk

Here's a statement I've come across multiple times but have never seen a proof of:

Two random variables $X$ and $Y$ are independent, if for all continuous and bounded funtions $f, g: mathbb Rtomathbb R$ it holds that
$$E[f(X)g(Y)]=E[f(X)]E[g(Y)].tag1$$

I found this answer but I'm not sure that I'm filling in the details correctly:

Suppose $X$ and $Y$ satisfy the condition in $(1).$ We want to show that $X$ and $Y$ are independent. Since the closed intervals generate the Borel sigma algebra, it suffices to show that
$$P(Xin I_1, Yin I_2)=P(Xin I_1)P(Yin I_2)$$
for all closed intervals $I_1, I_2subsetmathbb R.$ Given two such intervals let $f_n, g_nge0$ be sequences of continuous and bounded functions with
$$f_n(cdot)uparrow 1(cdotin I_1), quad g_n(cdot)uparrow 1(cdotin I_2). tag2$$

Then
beginalign*
P(Xin I_1, Yin I_2) &= E[1(Xin I_1, Yin I_2)]\
&= E[1(Xin I_1)1(Yin I_2)]\
&= E[lim_ntoinftyf_n(X)lim_mtoinftyg_m(Y)]\
&= lim_ntoinftyE[f_n(X)lim_mtoinftyg_m(Y)]quad text(by monotone convergence)\
&= lim_ntoinftylim_mtoinftyE[f_n(X)g_m(Y)]quad text(by m.c.)\
&= lim_ntoinftylim_mtoinftyE[f_n(X)]E[g_m(Y)]quad text(by assumption)\
&= E[1(Xin I_1)]E[1(Yin I_2)]quad text(by m.c.)\
&=P(Xin I_1)P(Yin I_2),\
endalign*
hence the claim.

Question: Is my above proof correct?

You cannot take closed intervals in this argument. The argument works if you take open intervals.