Let us consider a Lie algebra $\mathfrak g$ whose dimension is (at most) countable, i.e. it admits a countable set of generators $T_n$ with $n\in\mathbb N$, and let the linear space $V$ be an irreducible representation of $\mathfrak g$.
I would like to prove that if $\phi:V\to V$ is a homomorphism of representations of $\mathfrak g$, then $\phi = \lambda\, \mathrm{id}_V$ for some $\lambda \in \mathbb C$.
To this end, I was suggested to first show that $V$ may attain at most countable dimension. This first step can be carried out as follows: since $V$ is irreducible, given a reference vector $0\neq v \in V$, we may write any element $w\in V$ as a linear combination of vectors of the type $T_{i_1} T_{i_2}\cdots T_{i_n}v$; the latter form a countable basis, implying that $V$ has countable dimension as well.
At this point, I consider $\psi_\lambda \equiv \phi - \lambda\, \mathrm{id}_V$. Since $\phi$ is a homomorphism, so is $\psi_\lambda$, therefore both its kernel and its image are subrepresentations of $V$. But $V$ is irreducible, so $\psi_\lambda$ is either invertible (zero kernel and image equal to $V$) or identically zero (zero image, kernel equal to $V$) and in the latter case $\phi=\lambda\,\mathrm{id}_V$. The thesis is obtained once we conclude that $\psi_\lambda$ cannot be invertible for all $\lambda\in\mathbb C$.
I was shown a proof of this last step where one considers the polynomials in $\phi$ (for instance, see this Math.SE post). If one assumes that all such nonzero polynomials are invertible, then one coucludes that, for any set of $\lambda_k\in\mathbb C$, $$ \sum_{k} \frac{c_k}{\phi-\lambda_k\mathrm{id}_V}=0 $$ has only the trivial solution $c_k=0$, hence that all the $(\phi-\lambda\,\mathrm{id}_V)^{-1}$ are linearly independent; but this in contradiction with the fact that $V$ has at most countable dimension.
However, I was trying a more direct approach. Let us consider a fixed $0\neq v\in V$ and the set $\{\psi_\lambda(v): \lambda \in \mathbb C\}$. Since $V$ is countable dimensional, there are (infinitely many) nonzero solutions $c_k\neq 0$ to the equation $$ \sum_{k} c_k \psi_{\lambda_k}(v)=0\,, $$ that is $$ \sum_{k} c_k \phi(v) = \sum_{k}c_k \lambda_k v\,. $$ Now, if I were able to argue that the $c_k$ may be chosen such that $\sum c_k\neq 0$, I could conclude that $$ \phi(v) = \frac{\sum_{k}c_k \lambda_k}{\sum_{k} c_k} v\,. $$ This would indeed prove the theorem with $\lambda = (\sum c_k \lambda_k)/(\sum c_k)$. However every attempt at showing that $\sum c_k\neq 0$, for some choice, has proven unsuccessful until now.
Is there a way to make this type of argument right?