You can actually stretch the arguments to include the case of countable dimensional vector spaces if you are willing to work over $\mathbb C$:
Let $G$ be an abelian group and $\mathsf k$ an algebraically closed field. Write $\mathsf k[G]$ for the group algebra of $G$ over $\mathsf k$, that is,
$$
\mathsf k[G]= \{f\colon G\to \mathsf k: f(g)=0 \text{ for all but finitely many } g \in G\} = \text{span}_{\mathsf k}\{1_g: g\in G\}
$$
where, if $g,h \in G$ then $1_g(h) = 1$ if $h=g$ and $0$ otherwise, and $\mathsf k[G]$ has multiplication given by convolution, that is, $1_g.1_h = 1_{g.h}$.
Schur's Lemma:
If $(V,\rho)$ is an irreducible representation and $A=\text{Hom}_G(V,V)$ is the $\mathsf k$-algebra of $G$-endomorphism of $V$, then $A$ is a division algebra. That is, if $\alpha \in A$ then either $\alpha=0$ or $\alpha$ is invertible.
Proof
Let $\alpha\in A$ be non-zero. Because $\alpha$ commutes with the action of $G$, both $\text{ker}(\alpha)$ and $\alpha(V)$ are subspaces of $V$ which are preserved by $G$, and hence as $V$ is irreducible, we must have
- $\text{ker}(\alpha) = \{0_V\}$ or $V$, and
- $\alpha(V) = \{0_V\}$ or $V$.
But if $\text{ker}(\alpha)=V$ then $\alpha=0$, and so as $\alpha\neq 0$, $\text{ker}(\alpha)=\{0\}$. Similarly, if $\alpha(V)=\{0_V\}$ then again we would have $\alpha=0$, hence we must have $\alpha(V)=V$. Thus $\alpha$ is both injective and surjective and hence it is a bijection. Since the inverse of a linear map is automatically linear, $\alpha$ is invertible.
Now suppose that $\mathsf k$ is a field of uncountable cardinality. Then the field $\mathsf k(t)$ is of uncountable dimension over $\mathsf k$ (consider, for example, the elements $\{1/(t-\lambda): \lambda \in \mathsf k\}$).
Corollary:
Let $G$ be an abelian group and $(V,\rho)$ a representation of $G$ over an algebraically closed field $\mathsf k$ of uncountable cardinality (e.g., the complex numbers). If either
i). $V$ has countable dimension (i.e. a countable basis), or
ii). $G$ is countably generated (i.e. if $G$ has a countable set of generators)
then $\dim(V)=1$.
Proof:
First note that, for any group $G$ and representation $(V,\rho)$ of $G$, the homomorphism $\rho\colon G\to \text{GL}_{\mathsf k}(V)$ extends to a homomorphism of $\mathsf k$-algebras $\tilde{\rho}\colon \mathsf k[G]\to \text{End}_{\mathsf k}(V)$. When $G$ is abelian, we have $P = \tilde{\rho}(\mathsf k[G])\subseteq A$ and hence, since it is commutative, it is a field. Since $\mathsf k$ is algebraically closed, it follows that if $\alpha \in P$ then either $\alpha$ is transcendental over $\mathsf k$ or $\alpha \in \mathsf k.1_V$.
Thus either:
a) $P = \mathsf k.1_V$, or
b) there is some $g \in G$ with $\beta = \rho(g) \in A\backslash \mathsf k.1_V$, and hence we obtain an injective homomorphism $\tau_\beta\colon \mathsf k(t)\to P$ given by $f(t) = f(\beta)$ for all $f(t) \in \mathsf k(t)$.
Now, if $G$ is countably generated, then $\mathsf k[G]$ is a countably generated commutative algebra, and hence of countable dimension, and since it is a quotient of $\mathsf k[G]$, so is $P$. But then clearly $\mathsf k(t)$ cannot occur as a subfield of $P$, so case $ii)$ excludes $b)$ and hence implies $\dim(V)=1$.
Similarly, if $b)$ holds then we may also extend the structure of $V$ as a $\mathsf k$-representation of $G$ to that of a $\mathsf k(t)$-representation via $\tau_{\beta}$. But since $\dim_{\mathsf k}(\mathsf k(t))$ is uncountable, this contradicts the hypothesis $i)$, and hence in that case we must have $P=\mathsf k.1_V$ and so $V$ is $1$-dimensional.
Beyond the countable case, you almost surely need to use more than algebra. A good example (discussed above, but it worth noting) is $L^2(\mathbb R)$. This is a representation of the abelian group $(\mathbb R,+)$ which is not irreducible (it is not even indecomposable) but it contains no irreducible subrepresentations. Instead, the Fourier transform shows that it is a direct integral of irreducible representations $\mathbb C.\exp(i\lambda.x)$.
Additional remark: I am adding this in response to Severin Schraven's question in the comments about the existence of an infinite dimensional irreducible representation of an abelian group $G$. As others have pointed out, any irreducible representation of a finite group is necessarily finite-dimensional, so $G$ must be infinite for this to happen, but in fact we use an idea which also says something interesting in the finite-dimensional case when the field is not algebraically closed: If you want a representation of an abelian group over $\mathbb Q$ which is irreducible but has dimension greater than $1$, one way to do this is to take $$
\Phi_n = \prod_{\substack{1\leq d\leq n\\ (d,n)=1}} (t-\exp(2d\pi i/n)),
$$
the minimal polynomial of $\exp(2\pi i/n)$ over $\mathbb Q$, and set $F = \mathbb Q[t]/\langle \Phi_n(t)\rangle$. This is a field of degree $\phi(n)$ (where $\phi(n)$ is Euler's totient function) and the image $\alpha$ of $t$ is an primitive $n$-th root of unity in $F$. Thus $F$ yields a representation $(V,\rho)$ of the cyclic group $C_n$ on $V$ which is just $F$ viewed as a $\mathbb Q$ vector space. It is easy to see that $V$ is irreducible as a representation of $C_n$ since $F=\mathbb Q[\alpha]$.
Example of an infinite dimensional irreducible representation: Now let $\mathsf k$ be a field, and let $\mathsf F=\mathsf k(t)$ be the field of rational functions in $\mathsf k$. The $\mathsf F$ has dimension at least the cardinality of $\mathsf k$ over $\mathsf k$ (as one can see by considering $\{1/(t-\lambda): \lambda \in \mathsf k\}$. If we let $G$ be the abelian group $\mathsf F^*$ and $V$ be $\mathsf K$ viewed as a $\mathsf k$-vector space via the inclusion of fields $\mathsf k \hookrightarrow \mathsf F$, then $(V,\rho)$ is an infinite-dimensional $\mathsf k$-representation of $G$ over $\mathsf k$ which is irreducible: if $v \in V$ is non-zero, then the orbit $G.v = \mathsf F^*$ which certainly spans $\mathsf F$ over $\mathsf k$.
The above construction shows that we can make "big" irreducible representations of abelian groups by making $G$ "big", but this is also a little less esoteric than it sounds:
Fact: Any two algebraically closed fields of the same transcendence degree over $\mathbb Q$ are isomorphic.
Now $\mathbb C$ has uncountable transcendence degree over $\mathbb Q$, and it is not too hard to show that $\mathsf K$, the algebraic closure of $\mathbb C(t)$ also has uncountable transcendence degree thus as abstract fields $\mathbb C$ and $\mathsf K$ are isomorphic.
Using this fact, it follows that the group $\mathbb C^*$ has infinite-dimensional irreducible representations. Indeed take $\mathsf k=\mathbb C$ in the example above, and fix an isomorphism of fields $\tau\colon \mathbb C\to \mathsf K$. Then $(V,\rho\circ \tau_{|\mathbb C^*})$ is a representation of $\mathbb C^*$,
and since $\tau$ restricts to an isomorphism of groups $\mathbb C^* \cong \mathsf K^*$, the representation $(V,\rho\circ \tau_{|\mathbb C^*})$ is irreducible.
