Subgroups of $\Bbb{R}^n$ that are closed and discrete

Let $H\subset \mathbb R^n$ be such a subgroup. Note that no element of $H$ has finite order. Suppose that there exist elements $x_1,\ldots,x_{n+1}\in H$ which are linearly independent over $\mathbb Z$. They are also linearly independent over $\mathbb Q$ as clearing denominators shows, and over $\mathbb R$ since any real dependence relation can be approximated by rational ones, making $0$ an accumulation point. Then $H/\mathbb Zx_1\oplus\cdots \oplus \mathbb Zx_n$ is a subgroup of the $n$-torus $\mathbb R^n/\mathbb Zx_1\oplus\cdots \oplus \mathbb Zx_n$, and the image $\overline{x_{n+1}}$ of $x_{n+1}$ under the quotient map has infinite order. By compactness of the $n$-torus, the set $\mathbb Z\overline{x_{n+1}}$ has an accumulation point. Thus the set $\mathbb Zx_{n+1}$ comes arbitrarily close to the lattice $\mathbb Zx_1\oplus\cdots \oplus \mathbb Zx_n$, so $0$ is an accumulation point of $H$, a contradiction. Thus $H$ is generated by at most $n$ elements. The conclusion follows by applying the fundamental theorem of finitely generated abelian groups.

I fully endorse Alex Becker's answer. I just want to address the question of discreteness of $p(H)$ in the original post.

Let $W$ be the orthogonal complement to $L$, so $p$ is the orthogonal projection onto $W$.

Lemma. For all elements $y\in \left(H\setminus\mathbb{Z}x\right)$ we have $||p(y)||\ge ||x||/2.$

Proof. Let $y\in \left(H\setminus\mathbb{Z}x\right)$ be arbitrary. Let us write $$ y=rx+w, $$ where $r\in\mathbb{R}$ and $w\in W$. Because $x$ generates $L\cap H$, we must have $y\notin L$, so $w\neq0$. There exists an integer $m$ such that $|m-r|\le1/2$. Consider the vector $y'=y-mx=(r-m)x+w\in H$. Because $y\notin L$, $y'\neq0$. Therefore $||y'||\ge ||x||$ as $x$ was selected to be (one of) the shortest non-zero vectors in $H$. By construction $||(r-m)x||\le ||x||/2$. By triangle inequality $$ ||p(y)||=||w||=||y'-(r-m)x||\ge ||y'||-||(r-m)x||\ge ||x||-\frac{||x||}2=\frac{||x||}2. $$ Q.E.D.

The discreteness of the image follows immediately from the Lemma. If $p(y)$ and $p(y')$ are two distinct points in the group $p(H)$, then $y-y'\notin\mathbb{Z}x$, and therefore $$ d(p(y),p(y'))=||p(y)-p(y')||=||p(y-y')||\ge\frac{||x||}2. $$