I have proved the following statement, but I am not yet sure that everything is correct so I would appreciate comments and corrections, thanks.
$|A|=\lim_{t\to\infty} |A\cap (-t,t)|,\ A\subset\mathbb{R}, t>0$
NOTE: $|\cdot|$ refers to outer measure, i.e. for $A\subset\mathbb{R},\ |A|:=\inf\{\sum_{k=1}^{\infty}l(I_k): I_1,I_2,\dots\text{ are open intervals such that }A\subset\bigcup_{k=1}^{\infty}I_k\}$; the length of an open interval $I\subset\mathbb{R}$ is defined as
$l(I):=\begin{cases}
b-a & \text{if }I=(a,b),\ a,b\in\mathbb{R}, a<b; \\
0 & \text{if }I=\emptyset; \\
\infty & \text{if } I=(-\infty, a)\text{ or } I=(a,\infty);\\
\infty & \text{ if }I=(-\infty,\infty)
\end{cases} $
My proof:
We study the two possible cases: $|A|=\infty$ and $|A|<\infty$.
CASE $|A|=\infty$| Let $I_1,I_2,\dots$ be a sequence of open intervals such that $A\subset\bigcup_{k=1}^{\infty}I_k$. Then $\sum_{k=1}^{\infty}l(I_k)=\infty$ (since $|A|\leq\sum_{k=1}^{\infty}l(I_k)$) so if we take $M>0$ arbitrary there must be $K\geq 1$ such that $\sum_{k=1}^{K}l(I_k)>M$. We note that $I_k\cap(-t,t)$ is an open interval and $$A\cap(-t,t)\subset\bigcup_{k=1}^{\infty}(I_k\cap(-t,t))$$ so $$|A\cap(-t,t)|:=\inf\{\sum_{k=1}^{\infty}l(U_k):U_1,U_2,\dots\text{ open intervals such that }A\cap(-t,t)\subset\bigcup_{k=1}^{\infty}U_k\}=\inf\{\sum_{k=1}^{\infty}l(I_k\cap(-t,t)):I_1,I_2,\dots \text{ open intervals such that }A\subset\bigcup_{k=1}^{\infty}I_k\}$$ Now, if one the the $I_k$ (say, $I_{k*}$) is of the form $(-\infty,a),(a,\infty)$ ($a\in\mathbb{R}$) or $(-\infty,+\infty)$ we have that, in the first case $$I_k\cap(-t,t)=(-t,a)\Rightarrow l(I_k\cap(-t,t))=a-(-t)=a+t>M$$ for $t>\max\{M-a,a\}$, in the second case $$I_k\cap(-t,t)=(a,t)\Rightarrow l(I_k\cap(-t,t))=t-a>M$$ for $t>M+a$ and in the third case $$I_k\cap(-t,t)=(-t,t)\Rightarrow l(I_k\cap(-t,t))=2t>M$$ for $t>\frac{M}{2}$ thus in this case it's enough to take $t>\max\{M-a,M+a,\frac{M}{2}\}$ to have $\sum_{k=1}^{\infty}l(I_k\cap(-t,t))\geq l(I_{k*})>M$. If the $I_k$ are all finite ($I_k=(a_k,b_k)$) it's enough to take $t>\max\{|a_1|,|b_1|,|a_2|,|b_2|,\dots, |a_K|,|b_K|\}$ to have $\sum_{k=1}^{\infty}l(I_k\cap(-t,t))\geq\sum_{k=1}^{K}l(I_k\cap(-t,t))>M$. Thus we can conclude that if $t>\max\{|a_1|,|b_1|,\dots, |a_K|,|b_K|, M-a,M+a,\frac{M}{2}\}$ we have $\sum_{k=1}^{\infty}l(I_k\cap(-t,t))>M$ and taking the $\inf$ we get $|A\cap (-t,t)|>M$. Since $M>0$ was chosen arbitrarily we can conclude that for all $M>0$ there exists $t\geq 0$ such that $|A\cap (-t,t)|>M$ i.e. $\lim_{t\to\infty} |A\cap (-t,t)|=\infty=|A|$, as desired.
CASE $|A|<\infty|$ since $|A|=|A\cap (-t,t)|+|A\cap\mathbb{R}-(-t,t)|$ it is enough to show tha $\lim_{t\to\infty}|A\cap\mathbb{R}-(-t,t)|=\lim_{t\to\infty}|A\cap (-\infty,-t]\cup A\cap [t,+\infty)|=0$.
So, let $I_1,I_2,\dots$ be a sequence of open intervals ($I_n =(a_n, b_n)$ whose union contains $A$ (note that there cannot be an infinite one like $I_{n*}=(-\infty,a)$, $(a,\infty)$ or $(-\infty,\infty)$ since in that case $\sum_{n=1}^{\infty}l(I_n)\geq l(I_{n*})=\infty$).
For each $n\geq 1$ let $J_n:=I_n\cap (-\infty,-t]$ and $L_n:=I_n\cap [t,\infty)$. Since $A\cap (-\infty,-t]\cup A\cap [t,+\infty)\subset \bigcup_{n=1}^{\infty} (J_n\cup L_n)$ we have $$|A\cap (-\infty,-t]\cup A\cap [t,+\infty)|\leq |\bigcup_{n=1}^{\infty}(J_n\cup L_n)|\leq\sum_{n=1}^{\infty}(l(J_n)+l(L_n))=\sum_{n=1}^{\infty}l(J_n)+\sum_{n=1}^{\infty}l(L_n)$$
Now, let $\varepsilon >0$: there must be some $N\geq 1$ such that $|\sum_{n=1}^{\infty}l(J_n)-\sum_{n=1}^{N}l(J_n)|=|\sum_{n=N+1}^{\infty}l(J_n)|<\varepsilon/2$ and similarly there must be $N'\geq 1$ such that $|\sum_{n=1}^{\infty}l(L_n)-\sum_{n=1}^{N'}l(L_n)|=|\sum_{n=N'+1}^{\infty}l(L_n)|<\varepsilon/2$ so if $t>\max\{|a_{1}|,\dots,|a_{N}|,|b_{1}|,\dots, |b_{N'}|\}$ we have that $\sum_{n=1}^{\infty}l(J_n)+\sum_{n=1}^{\infty}l(L_n)<\varepsilon\Rightarrow |A\cap (-\infty,-t]\cup A\cap [t,+\infty)|<\varepsilon$. Thus, $\lim_{t\to\infty}|A\cap (-\infty,-t]\cup A\cap [t,+\infty)|=0$, as desired.
Best Answer
To me it seems that your argument works out if $A$ is bounded. To avoid infinite sums, you may then take a finite covering of $A$ with open intervals of radius $\epsilon>0$ (since the closure of $A$ is compact). In other words, you can take the sequence $(a_n,b_n)$ such that $b_n-a_n=\epsilon>0$ for any $\epsilon>0$ and finitely many such intervals suffice to cover $A$.
I'm not convinced that your argument works directly for unbounded sets. I think the partial sums are not guaranteed to be bounded by $\epsilon/2$.
Here is a similar approach which seems to work generally. The idea is the same, but I choose the intervals $(a_n,b_n)$ carefully, to benefit from the result you provided. Since you know: $$|A|=|A\cap (-t,t)|+|A\setminus (-t,t)|$$ It also holds that for $t'>t$: $$ |A\setminus (-t,t)|=|A\setminus (-t,t)\cap (-t',t')|+|A\setminus (-t,t)\setminus (-t',t')| $$ $$ =|A\cap((-t',-t]\cup[t,t'))|+|A\setminus (-t',t')| $$ So, we may write: $$ |A|=\sum_{n=0}^\infty|A\cap ((-n-1,-n]\cup[n,n+1))| $$ To see that the sum indeed converges to $|A|$, note that if $|A|=+\infty$, the sum must diverge and if $|A|<+\infty$, then the sum converges monotonely to $|A|$. Then, you have for $t>0$: $$ \sum_{n=0}^{\lfloor t\rfloor}|A\cap ((-n-1,-n]\cup[n,n+1))|\leq |A\cap (-t,t)| $$ $$ \leq \sum_{n=0}^{\lceil t\rceil}|A\cap ((-n-1,-n]\cup[n,n+1))| $$ As $t\rightarrow\infty$, both the left and right side converge monotonely to $|A|$, hence so does $ |A\cap (-t,t)|$.