I am well aware that there is an answer to this question. In fact there is an answer here.
Question 7.7 in measure theory on Radon measure from Folland's Real Analysis Second Edition
This solutions seems quite long, much longer than my solution. I solved this in a relatively very simple fashion which is leading me to believe there is something wrong with my solution
The question from Folland is
If $\mu$ is a $\sigma$-finite radon measure on $X$ and $A\in \mathcal{B}_{X}$, the Borel measure $\mu_{A}$ defined by $\mu_{A}(E)=\mu(E\cap A)$ is a radon measure.
My solution:
-It should first be noted that since $\mu$ is $\sigma$-finite, it follows that it is a regular measure by Corollary 7.6. That is, it is inner regular on all Borel sets, not just open sets.
-For $K$ compact we clearly have $\mu_{A}(K)=\mu(K\cap A)\leq \mu(K)<\infty$ since we know $\mu$ is radon
-Outer regularity: Let $E\in \mathcal{B}_{X}$. Then since $\mu$ is sigma finite we have from proposition 7.7 that for $\epsilon>0$ given there is open $V$ such that $\mu(V-E)<\epsilon$. Note that this is true even if $\mu(E)=\infty$. It then follows that $\mu_{A}(V-E)=\mu((V-E)\cap A)\leq \mu(V-E)<\epsilon$. Thus outer regularity has been proved.
-Inner regularity: If $\mu(A\cap V) < \infty$ then using inner regularity of $\mu$ there is a compact $K\subset A\cap V$ such that $\mu(A\cap V-K) <\epsilon$. Thus $\mu_{A}(V-K)=\mu((V-K)\cap A)=\mu(V\cap A-K\cap A)=\mu(V\cap A-K)<\epsilon$
If $\mu(A\cap V)=\infty$. Then write $A=\cup_{i=1}^{\infty}A_{i}$ with the $A_{i}$ mutually disjoint and $\mu(A_{i})<\infty$. Then using inner regularity of $\mu$, choose $K_{i}\subset A_{i}\cap V$ such that $\mu(A_{i}\cap V-K_{i})<\frac{\epsilon}{2^i}$. Then $\mu_{A}(V-\cup_{i=1}^{\infty}K_{i})=\mu(\cup_{i=1}^{\infty}(A_{i}\cap V-K_{i}))=\sum_{i=1}^{\infty}\mu(A_{i}\cap V-K_{i})<\epsilon$. Since $\mu_{A}(V)=\infty$, this implies that $\mu_{A}(\cup_{i=1}^{\infty}K_{i})=\infty$. Thus $\mu_{A}(\cup_{i=1}^{n}K_{i})\rightarrow \infty$ and $n\rightarrow \infty$ and $\cup_{i=1}^{\infty}K_{i}\subset V$. Thus $\mu_{A}$ is inner regular in the case $\mu_{A}(V)=\infty$ as well.
In particular, the linked solution seems to do a lot more work for outer regularity which leads me to believe my solution has an error somewhere. I have read the linked solutions and can't understand why they are doing so much work for outer regularity.
If someone could point out the error in my reasoning then that would be great.
Your proof is fine, assuming that you have already proved proposition 7.7, which is where all the hard work is. In the linked proof the poster is basically proving proposition 7.7 part a), which is why it is longer. You'll note that the linked post proves what you use from proposition 7.7 for showing both inner and outer regularity.