= Solution
<Schur lemma> says that an intertwining linear map between irreducible finite-dimensional complex representations is either zero or an isomorphism. In particular, every endomorphism of an irreducible complex representation is a scalar multiple of the identity.
A <continuous representation of a topological group> is a continuous homomorphism
$$
\rho:G\longrightarrow\operatorname{GL}(V)
$$
for a finite-dimensional complex <vector space> $V$. It is a <unitary representation> if $V$ has a positive-definite Hermitian <inner product> for which
$$
\langle\rho(g)v,\rho(g)w\rangle
=\langle v,w\rangle
$$
for every $g\in G$ and $v,w\in V$.
For $G=S^1$, start with any positive-definite Hermitian form and average it using normalized <Haar measure>:
$$
\langle v,w\rangle_{S^1}
=\int_{S^1}
\langle\rho(z)v,\rho(z)w\rangle\,dz.
$$
Translation invariance makes this form $S^1$-invariant, and positivity is preserved, proving <unitarization of a compact-group representation>[unitarity]. Since $S^1$ is abelian, the operators $\rho(z)$ commute; since they are unitary, they are <normal matrix>[normal]. <Simultaneous diagonalization> therefore decomposes $V$ into common one-dimensional eigenspaces. Thus every <representation of the circle group> is a direct sum of one-dimensional representations.
Write
$$
g(x,y,z)=
\begin{pmatrix}
1&x&z\\0&1&y\\0&0&1
\end{pmatrix}.
$$
The group law and inverse are
$$
\begin{aligned}
g(x,y,z)g(x',y',z')
&=g(x+x',y+y',z+z'+xy'),\\
g(x,y,z)^{-1}
&=g(-x,-y,-z+xy).
\end{aligned}
$$
A calculation gives
$$
g(x,y,z)^{-1}g(x',y',z')^{-1}g(x,y,z)g(x',y',z')
=g(0,0,xy'-x'y).
$$
Every element of the <center of a group>[centre] $Z$ occurs by taking, for example, $x'=y=0$ and $y'=1$. Hence the <commutator subgroup> is exactly $Z$. The image of a one-dimensional representation is abelian, so the <one-dimensional representation kills the commutator subgroup> and its kernel contains $Z$.
Now let $(\rho,V)$ be a complex representation of $G/Z_0$. Its restriction to the central subgroup
$$
Z/Z_0\cong\mathbb R/\mathbb Z\cong S^1
$$
is a representation of the circle group. Decompose it into its distinct weight spaces:
$$
V=V_1\oplus\cdots\oplus V_d,
\qquad
\rho(z)|_{V_i}=\theta_i(z)\operatorname{id}_{V_i}.
$$
Because $Z/Z_0$ is central, every $\rho(g)$ commutes with its action and preserves every common eigenspace. Thus the $V_i$ are $G/Z_0$-subrepresentations, as asserted by the <central circle weight-space decomposition>.
Let $n_i=\dim V_i$. The map
$$
g\longmapsto\det(\rho(g)|_{V_i})
$$
is a one-dimensional representation of $G/Z_0$; after composition with $G\to G/Z_0$, its kernel contains $Z$. On $z\in Z/Z_0$ its value is
$$
\det(\theta_i(z)I_{n_i})=\theta_i(z)^{n_i},
$$
so $\theta_i^{n_i}=1$. A continuous character of $S^1$ has the form $z\mapsto z^m$; the displayed identity forces $m=0$. Therefore every $\theta_i$ is trivial. Since the $\theta_i$ were distinct, $d=1$ and $\theta_1=1$.
It follows that every finite-dimensional complex representation of $G/Z_0$ kills the entire nontrivial central circle $Z/Z_0$. Its kernel is therefore nontrivial, so no such representation is <faithful representation>[faithful]. This is the <real Heisenberg quotient has no faithful finite-dimensional representation>.
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