Let $V$ be an infinite-dimensional $K$-vector space, $E=\operatorname{End}_K(V)$ and $I = \{f \in E:\dim_Kf(V) < \infty\}$.
I want to prove that $I$ is an ideal of $E$ and $K\mathrm{id}_V + I$ is a subring of $E$.
My idea:
Let's first prove that $I$ is an ideal of $E$. Let $f$ be an element of $I$ and $g$ an element of $E$.
Then, $dim((f \circ g)(V)) \le dim(f(g(V))) \le dim(f(V)) \lt \infty$ and $dim((g \circ f)(V)) \le dim(g(f(V))) \lt \infty$. This proves that $I$ is an ideal of $E$.
Let' s prove next that $Kid_V + I$ is a subring of $E$. Let $k_1id_V + f_1$ and $k_2id_V+f_2$ be elements of $I$.
Then, $(k_1id_V+f_1) \circ (k_2id_V+f_2) = k_1k_2id_V + k_1f_2 + k_2f_1 + f_1 \circ f_2$.
Futhermore, $dim(k_1f_2 + k_2f_1 + f_1 \circ f_2)\le dim(k_1f_2) + dim(k_2f_1) + dim(f_1 \circ f_2) \lt \infty$.
Besides, $id_V \in I$ and $-(k_1id_V+f_1) \in I$ as well as $0 \in I$.
This proves that $I$ is a subring.
Is this correct?
By the way, does someone know if $V$ is a simple module?
You needn't be checking dimensions when considering $Kid_V+I$. You say $id_V\in I$, but this is obviously false: the image of $id_V$ is all of $V$, which is infinite dimensional by definition. This new ring properly contains $I$, and so it necessarily contains things with infinite dimensional images.
It is enough to show that the set of things of the form $\{k\cdot1_V+ i\mid k\in K, i\in I\}$ is closed under addition and multiplication, and that is easy enough. The only thing in contention really is whether or not $(k\cdot 1_V)\cdot i$ and $i\cdot (k\cdot 1_V)$ have finite dimensional images, but knowing that $I$ is an ideal makes this obvious.
Yes, $V$ is a simple module (when acted upon by linear transformations from this ring in the obvious way.) Given $v,w$ nonzero elements of $V$, you can easily construct a linear transformation that takes one to the other. This amounts to the module being simple.