Dixmier's lemma as a generalisation of Schur's first lemma

If $M-c$ is not invertible, then its kernel is non-zero or its image is smaller than $A$. But $A$ is irreducible. This means any proper submodule is zero. This forces $M-c=0$.

BTW, for finite-dimensional Lie algebras it works over $\overline{\mathbb{Q}}$ as well. It is known as Quillen Lemma. A good discussion when it works can be found in Noncommutative Noetherian Rings by McConnell and Robson.

Yes, it indeed is correct that Dixmier's lemma and Schur's lemma, together, entail the stronger statement that $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,=\,0\,$ and, therefore, $\,{\mathbb{M}}\,$ is proportional to the identity operator.

See Lemma 99 in https://arxiv.org/abs/1212.2578 , where that story follows N. R. Wallach's "Real Reductive Groups. 1."

I'm writing this just to give context, as others already answered the question. In the discussion below we fix a field $k$ of cardinality $|k|$.

The following version of Schur's lemma is very general, also trivial to prove: For an irreducible $k$-representation of a group $G$ the $k$-algebra of intertwiners is a $k$-division algebra.

By a $k$-division algebra we mean, as usual, a unital $k$-algebra in which every non-zero element is invertible and $k$ is central in it.

In fact, every $k$-division algebra $D$ could be seen as the algebra of intertwiners of an irreducible $k$-representation for some group $G$ - just take $G=D^*$ and consider its left regular representation on $D$.

Note that $\dim k(x)\geq |k|$, as the set $\{(x-\alpha)^{-1}\mid \alpha\in k\}$ is linearly independent (in fact an equality holds for infinite $k$). By fixing $d\in D-k$ and considering the evaluation map $k(x)\to D$, $x\mapsto d$ we get

Lemma: If $k$ is algebraically closed and $D$ is a non-trivial $k$-division algebra then $\dim(D)\geq \dim k(x) \geq |k|$.

If $G\to \text{GL}(V)$ is an irreducible representation with $D=\text{End}(V)^G$ then, fixing $0\neq v\in V$, the map $D\to V$, $d\mapsto d(v)$ is clearly injective, thus $\dim(D)\leq\dim(V)$.

It follows from the discussion above that if $k$ is algebraically closed and $\dim(V)<|k|$ then $\text{End}(V)^G=k$ (and the cardinality condition is strict by considering the regular representation of $k(X)$).