possible duplicate but want work checked and verified
let $l =$ any even number and $m =$ any even number
$l+m = w$ where $w$ is any element in the even integers thus it is closed under addition
$2k+0=2k$ and $0-2k=2k$ so there is an identity both left and right
every even number can have its negative added to get 0 so there is an inverse for every element in the even integers i.e $2+(-2)=0$
Its associative
On
Your proof is fine.
Here's a more general result.
Theorem: For $k\in \Bbb Z,$ we have $$k\Bbb Z\le \Bbb Z.$$
Proof: Let $k\in \Bbb Z$. We use the one-step subgroup test; that is, we show that $\emptyset\neq k\Bbb Z\subseteq \Bbb Z$ and for any $m,n\in k\Bbb Z$, written additively, $m-n\in \Bbb Z$.
Indeed, $0=k0\in k\Bbb Z$, so $k\Bbb Z$ is nonempty. Let $\ell\in k\Bbb Z$. Then $\ell=kt$ for some $t\in \Bbb Z$, so $\ell$ is an integer as $k, t\in\Bbb Z$. Thus $k\Bbb Z\subseteq \Bbb Z$.
Let $m, n\in k\Bbb Z$. Then $m=ka, n=kb$ for some $a, b\in\Bbb Z$. But now $m-n=ka-kb=k(a-b)$ and $a-b\in\Bbb Z$. Hence $m-n\in k\Bbb Z$.
Hence $k\Bbb Z \le \Bbb Z$. $\square$
Now just let $k=2$ in the theorem above.
Your proof is okay, but it can be improved.
Closed: For any $l,m \in \mathbb{E}$ (where $\mathbb{E}$ is the set of even integers) $l = 2a$ and $m = 2b$ for some $a,b \in \mathbb{Z}$. Thus, $l + m = 2a + 2b = 2(a+b) \in \mathbb{E}$.
Identity: $0 \in \mathbb{E}$ and $0 + m = m + 0 = m$ for all $m \in \mathbb{E}$.
Inverse: For any $2k \in \mathbb{E}$, the number $2(-k) \in \mathbb{E}$ and $2k + 2(-k) = 2k - 2k = -2k + 2k = 0$.
Associative: The integers under addition are associative, so $\mathbb{E} \subset \mathbb{Z}$ inherits that property.