Does the Helmholtz-Leray projection map $H^1$ to itself?

Let

• $d\in\mathbb N$
• $\Lambda\subseteq\mathbb R^d$ be open
• $\mathcal V:=\left\{\phi\in C_c^\infty(\Lambda,\mathbb R^d):\nabla\cdot\phi=0\right\}$ and $$H:=\overline{\mathcal V}^{\left\|\;\cdot\;\right\|_{L^2(\Lambda,\:\mathbb R^d)}}$$
• $\operatorname P_H$ denote the orthogonal projection from $L^2(\Lambda,\mathbb R^d)$ onto $H$

How can we show that $\text P_HH^1(\Lambda,\mathbb R^d)\subseteq H^1(\Lambda,\mathbb R^d)$?

There is an outline of a proof in Remark 1.6 of Navier-Stokes Equations: Theory and Numerical Analysis by Roger Temam. However, I don't understand the following conclusion:

Let $$G:=\left\{\nabla p:p\in L_{\text{loc}}^2(\Lambda)\text{ has a weak gradient }\nabla p\in L^2(\Lambda,\mathbb R^d)\right\}\;.$$ Then $$H=G^\perp=\left\{u\in L^2(\Lambda,\mathbb R^d):\langle u,\nabla p\rangle_{L^2(\Lambda,\:\mathbb R^d)}=0\text{ for all }\nabla p\in G\right\}\;.\tag 1$$ Now, let $u\in H^1(\Lambda,\mathbb R^d)$. Then, $$u=v+\nabla p\tag 2$$ for some unique $(v,\nabla p)\in H\times G$. It's easy to verify that the distributional divergence $\nabla\cdot v$ of $v$ vanishes and hence $$\nabla\cdot u=\Delta p\;.\tag 3$$ By definition of $u$, $$\nabla\cdot u\in L^2(\Lambda)\tag 4$$ and hence the distributional Laplacian $\Delta p$ of $p$ is in $L^2(\Lambda)$, i.e. $\Delta p$ is the weak Laplacian defined by the relation $$\langle\phi,\nabla\cdot u\rangle_{L^2(\Lambda)}=-\langle\nabla\phi,\nabla p\rangle_{L^2(\Lambda,\:\mathbb R^d)}\;\;\;\text{for all }\phi\in C_c^\infty(\Lambda)\;.\tag 5$$

Now, Temam is concluding that $p\in H^2(\Lambda)$. I don't understand that and I don't think that $(5)$ implies that the distributional partial derivatives $\frac{\partial^2p}{\partial x_i\partial x_j}$ are in $L^2(\Lambda)$. So, what am I missing?