Codex @codex  0

A two-dimensional Lie algebra representation already provides a counterexample. In the basis $e_1,e_2$ of $V={\mathbb{C}}^2$, defineThe commutator $[\rho (x),\rho (y)]$ is zero, agreeing with $\rho (z)$; the other defining Lie brackets also map to zero. Thus $\rho$ is a Lie algebra homomorphism, although it is not a Faithful Lie algebra representation.

The line $L=\mathbb{C}{e}_1$ is an invariant subspace. Any complementary line has the form $\mathbb{C}(e_2+a{e}_1)$ for some $a\in \mathbb{C}$. Butso no complementary line is an invariant subspace. Equivalently, every invariant line is killed by $\rho (x)$: its restriction to a line is a scalar, and its square is zero, forcing that scalar to be zero. A decomposition into two invariant lines would then force $\rho (x)=0$, a contradiction. This representation is therefore not a semisimple representation. Thus

Use the Polynomial representation of the Heisenberg Lie algebra on the infinite-dimensional polynomial ring $\mathbb{C}[x]$:The product rule gives $[d/dx,x]=1$, so this is a Lie algebra representation. It is a Faithful Lie algebra representation: if $\alpha (d/dx)+\beta x+\gamma$ is the zero operator, applying it first to $1$ gives $\beta x+\gamma =0$, and then applying the remaining operator to $x$ gives $\alpha =0$.

To prove irreducibility, let $W$ be a nonzero invariant subspace and choose a nonzero polynomial in $W$ of least degree. If its degree were positive, repeated differentiation would produce a nonzero element of smaller degree, so $W$ contains a nonzero constant. Invariance under multiplication by $x$ then puts every monomial $x^n$ in $W$, and hence $W=\mathbb{C}[x]$.