*You are quite close to solving the problem:*

Using the representation $X_k=\sum_{j=1}^k Y_{kj}$ where $Y_{ij}\stackrel{\text{iid}}{\sim}\mathcal{P}(1)$, you have
$$\sum_{k=1}^n X_k= \sum_{k=1}^n\sum_{j=1}^k Y_{kj} = \sum_{u=1}^{n(n+1)/2} \xi_{u}$$where $\xi_u\stackrel{\text{iid}}{\sim}\mathcal{P}(1)$ $(u=1,\ldots,n(n+1)/2)$. Therefore, if you normalise the above sum, you get
$$\eqalign{\dfrac{\sum_{k=1}^n X_k-\mathbb{E}[\sum_{k=1}^n X_k]}{\text{var}(\sum_{k=1}^n X_k)^{1/2}}
&=\dfrac{\sum_{u=1}^{n(n+1)/2} \xi_{u}-\mathbb{E}[\sum_{u=1}^{n(n+1)/2} \xi_{u}]}{\text{var}(\sum_{u=1}^{n(n+1)/2} \xi_{u})^{1/2}}\\
&=\dfrac{\sum_{u=1}^{n(n+1)/2} \xi_{u}-\frac{n(n+1)}{2} }{(n(n+1)/2)^{1/2}}\\
&=\sqrt{2}\,\dfrac{\sum_{k=1}^n X_k-\frac{n^2}{2}-\frac{n}{2}}{n(1+n^{-1})^{1/2}}\\
&=\sqrt{2}\,\dfrac{\frac{1}{n}\sum_{k=1}^n \left[X_k-\frac{n^2}{2}\right]-\frac{1}{2}}{(1+n^{-1})^{1/2}}\\
&=\sqrt{2}\,\dfrac{Z_n-\frac{1}{2}}{(1+n^{-1})^{1/2}}\\
}$$
which should help you conclude, along with a CLT on the above. 

> The theoretical references for a Central Limit Theorem for independent
> _but not i.i.d._ random variables are [Liapounov][1]'s and [Lindeberg][2]'s versions of the CLT. The former requires moments of
> order $2+\epsilon$ with $\epsilon>0$ and the latter for [vanishing
> tail second moments][3]. Both conditions apply for Poisson variates.

That the result holds (and hence that there is no mistake in the formulation) can be checked by a quick R experiment, as illustrated by the following that compares an histogram of 10³ $Z_n$'s with the $\text{N}(1/2,1/2)$ density:

[![enter image description here][4]][4]


  [1]: https://en.wikipedia.org/wiki/Central_limit_theorem#Lyapunov_CLT
  [2]: https://en.wikipedia.org/wiki/Central_limit_theorem#Lindeberg_CLT
  [3]: https://en.wikipedia.org/wiki/Lindeberg's_condition
  [4]: https://i.sstatic.net/Y89kS.jpg