"Then the claim would be immediate by
semisimplicity if one can show that
every irreducible representation of
$G(\mathbb{C})$ (or perhaps of Lie
groups in some more general class than
these) occurs as a subquotient of a
tensor power of a faithful one. How
might one prove the latter? (Can one
prove the latter for compact real
groups in a manner similar to the
proof for finite groups, and then pass
to semisimple complex groups by the
unitary trick?)"
Yes. The analysis is not so pretty, but it is elementary. Let $K$ be a compact Lie group, $V$ a faithful representation, and $W$ any other representation. Just as in the finite group case, $\mathrm{Hom}_K(W,V^{\otimes N}) \cong (W^* \otimes V^{\otimes N})^K$, and the dimension of the latter is $\int_K \overline{\chi_W} \cdot \chi_V^{N}$, where $\chi_V$ and $\chi_W$ are the characters of $V$ and $W$, and the integral is with respect to Haar measure. Let $d_V$ and $d_W$ be the dimensions of $V$ and $W$.
We now come to a technical nuisance. Let $Z$ be those elements of $K$ which are diagonal scalars in their action on $V$; this is a closed subgroup of $S^1$. For $g$ not in $Z$, we have $|\chi_V(g)| < d_V$. We first present the proof in the setting that $Z = \{ e \}$.
Choose a neighborhood $U$ of $\{ e \}$ small enough to be identified with an open disc in $\mathbb{R}^{\dim K}$. On $U$, we have the Taylor expansion $\chi_V(g) = d_V \exp(- Q(g-e) + O(g-e)^3)$, where $Q$ is a positive definite quadratic form; we also have $\chi_W(g) = d_W + O(g-e)$. Manipulating $\int_U \overline{\chi_W} \chi_V^N$ should give you
$$\frac{d_W \pi^{\dim K/2}}{\det Q} \cdot d_V^N \cdot N^{-\dim K/2}(1+O(N^{-1/2}))$$
Meanwhile, there is some $D<d_V$ such that $|\chi_V(g)| < D$ for $g \in K \setminus U$.
So the integral of $\overline{\chi_W} \chi_V^N$ over $K \setminus U$ is $O(D^N)$, which is dominated by the $d_V^N$ term in the $U$ integral.
We deduce that, unless $d_W=0$, we have $\mathrm{Hom}_K(W, V^{\otimes N})$ nonzero for $N$ sufficiently large.
If $Z$ is greater than $\{e \}$, then we can decompose $W$ into $Z$-isotypic pieces. Let $\tau$ be the identity character of the scalar diagonal matrices, and let the action of $Z$ on $W$ be by $\tau^k$. (If $\tau$ is finite, then $k$ is only defined modulo $|Z|$; just fix some choice of $k$). Then we want to consider maps from $W$ to $V_N:=V^{k+Nd_V} (\det \ )^{-N}$. $V_N$ is constructed so that $\overline{\chi_W} \chi_{V_N}$ is identically $d_W$ on $Z$; one then uses the above argument with a neighborhood of $Z$ replacing a neighborhood of the origin.
Edit: I now give the argument for general reductive $G$.
Let $G$ be a reductive algebraic group over an alg. closed field $k$ of char. 0. Fix a max
torus $T$ and write $X = X^*(T)$ for its group of characters. Write $R$ for the
subgroup of $X$ generated by the roots of $G$. Then the center $Z$
of $G$ is the diagonalizable subgroup of $T$ whose character group is $X/R$.
Claim: $G$ has a faithful irreducible representation if and only if the character
group $X/R$ of $Z$ is cyclic.
Note for semisimple $G$, the center $Z$ is finite. Since the characteristic of $k$ is 0, in this case the group of $k$-points of $Z$ is (non-canonically) isomorphic
to $X/R$. Thus $Z$ is cyclic if and only $X/R$ is cyclic.
In general, the condition that $X/R$ is cyclic means either that the group of points $Z(k)$ is finite cyclic, or that $Z$ is a 1 dimensional torus.
As to the proof, for $(\implies)$ see Boyarsky's comment following reb's answer.
For $(\Leftarrow)$ let me first treat the case where $G$ is almost simple; i.e. where the root system $\Phi$ of $G$ is irreducible. Supopse that the class of $\lambda \in X$ generates the cyclic group $X/R$. Since
the Weyl group acts on $X$ leaving $R$ invariant, the class of any $W$-conjugate of $\lambda$ is also a generator of $X/R$. Thus we may as well suppose $\lambda$ to be dominant
and non-0 [if $X=R$, take e.g. $\lambda$ to be a dominant root...]
Now the simple $G$-module $L=L(\lambda) = H^0(\lambda)$ with "highest weight $\lambda$"
will be faithful. To see this, note that since $\lambda \ne 0$, $L$ is not the trivial representation. Since $G$ is almost simple, the only proper normal subgroups
are contained in $Z$. Thus it suffices to observe that the action of $Z$ on the
$\lambda$ weight space of $L$ is faithful.
The general case is more-or-less the same, but with a bit more book-keeping.
Write the root system $\Phi$ of $G$ as a disjoint union $\Phi = \cup \Phi_i$
of its irreducible components. There is an isogeny
$$\pi:\prod_i G_{i,sc} \times T \to G$$
where $T$ is a torus and $G_{i,sc}$ is the simply connected almost simple group
with root system $\Phi_i$. Write $G_i$ for the image $\pi(G_{i,sc}) \subset G$.
The key fact is this:
a representation $\rho:G \to \operatorname{GL}(V)$ has
$\ker \rho \subset Z$ if and only if the restriction $\rho_{\mid G_i}$ is non-trivial
for each $i$.
Now, as before pick $\lambda \in X$ for which the coset of $\lambda$ generates
the assumed-to-be cyclic group $X/R$. After replacing $\lambda$ by a Weyl group
conjugate, we may suppose $\lambda$ to be dominant. After possibly repeatedly replacing
$\lambda$ by $\lambda + \alpha$ for dominant roots $\alpha$, we may suppose
that $\lambda$ has the following property:
$$(*) \quad \text{for each $i$, there is $\beta_i \in \Phi_i$ with $\langle \lambda,\beta_i^\vee \rangle \ne 0$}$$
Now let $L = L(\lambda)$ be the simple module with highest weight $\lambda$. Condition
$(*)$ implies that $G_i$ acts nontrivially on $L$ for each $i$, so by the "key fact",
the kernel of the representation of $G$ on $L$ lies in $Z$. but since $\lambda$
generates the group of characters of $Z$, the center $Z$ acts faithfully on the
$\lambda$ weight space of $L$.
Best Answer
As requested by Faisal, I am posting as an answer the observation that if $G$ has more components than the size of the complex numbers then G has no faithful finite-dimensional irreducible representation over the complexes, for cardinality reasons.
To be honest I thought Faisal's response to this would be "what happens if $G$ has only countably infinitely many components"? That is then a different question. Does every countable group have a faithful finite-dimensional complex representation? If this is well-known I'd be happy to hear the answer. If it isn't I'd be happy if someone else asked this as a separate question. If it's true that any countable group has a faithful representation then one might ask about complex Lie groups with countably infinitely many components.