I'd just like to verify whether I've done it correctly.
Proof strategy. First we determine what we need to set $N$ equal to; $$\left|\frac{n+2}{2n+3}-\frac12\right|=\left|\frac{2n+4-(2n+3)}{4n+6}\right|= \frac1{4n+6}<\epsilon$$ Some rewriting get us, $$n>\frac{1}{4\epsilon}-\frac{6}4{}$$ When $\epsilon$ is $\frac{1}{6}$ $N$ will be 0 but this is not allowed since N should be a positive integer. We notice: $$n>\frac1{4\epsilon}>\frac1{4\epsilon} -\frac64$$ Hence choose $N=\lceil\frac1{4\epsilon}\rceil$
Proof: Let $\epsilon>0$ and choose $N=\lceil1/4\epsilon\rceil$. Let $n>N$ where $n \in \mathbb{Z}$. Then $n>\frac1{4\epsilon}>\frac1{4\epsilon}-\frac64$ .
Thus, rewrite to get, $ \frac{1}{4n+6} < \epsilon$
Therefore, $$\left|\frac{n+2}{2n+3}-\frac12\right|=\left|\frac{2n+4-(2n+3)}{4n+6}\right|=\frac1{4n+6}<\epsilon$$
Hence the sequence converges to $\frac{1}{2}$. $$\space \blacksquare$$
Yes, it is correct, apart from the part where you write "We notice n > ...". You should omit the part "n > " there and simply write "We notice $\frac1{4\varepsilon} > \frac1{4\varepsilon} - \frac64.$"