Would this solution of the limit of the sequence be correct?

Your solution is fine. Here is an other way to get the result. I know it is not the easier way to solve it, but the advantage is that you can generalize to $b_n = \sum_{k=1}^{n} \frac{k^2}{n^3}$ for instance.

Notice that $$a_n = \frac{1}{n} \sum_{k=0}^{n} \frac{k}{n}$$ looks like a Riemann sum for $f : x \mapsto x$ since $$a_n = \sum_{k=0}^{n} f\left(\frac{k}{n}\right) \left(\frac{k+1}{n} - \frac{k}{n} \right)$$

Therefore the limit of the sequence $(a_n)_{n≥1}$ is $$\int_0^1 f = \left[\frac{x^2}{2}\right]_0^1 = \frac{1}{2}$$

This seems correct. Personally I would not say $\lim\frac {n+1}{2n}=\lim\frac n{2n}$. Not without explanation at least (e.g. include the step $\lim \frac{n+1}{2n}=\lim \left(\frac{n}{2n}+\frac 1{2n}\right)$). I would rather say $\lim \frac {n+1}{2n}=\lim \frac{1+\frac1n}{2}$. But this is of course a matter of preference.