It is a remarkable and beautiful fact that the irreducible representations of the symmetric group $S_n$ are in correspondence with the partitions of $\lambda \vdash n$. For example, in the case of $S_3$, the irreducible partitions correspond to all the partitions of 3, namely
$$(3) \quad (1,1,1) \quad (2,1)$$
The full story is too long for this post, but details can be found in Chapter 4 of 'Representation Theory: A first course' by Fulton and Harris.
To see the decomposition of the regular representation into its irreducible components is most easily done via character theory. Let $\chi$ be the character of the left regular representation (that is, let $S_3$ act on $K[S_3]$ on the left); $\chi(\sigma)$ is the number of fixed points of the action of $\sigma$ on $K[S_3]$ (as these contribute to the trace of this action). It is plain to see that the only element in $S_3$ which fixes anything in $K[S_3]$ is the identity element $e$, and moreover, this fixes every element in $K[S_3]$. Thus we have
$$\chi(e) = |S_3|=6 \qquad \chi(\sigma)=0 \quad \forall \ \sigma \in S_3 \backslash \{e\}$$
Let $\chi_\lambda$ be the character of the irreducible representation of $S_3$ corresponding to the partition $\lambda \vdash 3$. It is a fact from character theory that the inner product of characters of a representation $A$ and an irreducible representation $B$, defined by,
$$\langle \chi_A, \chi_B \rangle = \frac{1}{|G|}\left( \sum_{g \in G} \chi_A(g)\chi_B(g) \right)$$
gives the multiplicity of $B$ in $A$. For your question then, we need to compute this inner product with the character $\chi_\lambda$. For this we only need to know one more fact: that $\chi_\lambda(e)$ is the dimension of the corresponding irreducible representation of $S_3$ corresponding to $\lambda$. We can now compute
$$\langle \chi, \chi_\lambda \rangle = \frac{1}{|S_3|}\left(\chi(e)\chi_\lambda(e) \right) = \frac{1}{6}(6\cdot \chi_\lambda(e)) = \chi_\lambda(e)$$
We see that the irreducible representation corresponding to $\lambda$ appears in the decomposition of the regular representation exactly the 'dimension of representation' number of times.
Here are the correspondences in your case:
$$\lambda = (3) \rightarrow V_0, \dim = 1$$
$$\lambda = (1,1,1) \rightarrow V_1, \dim = 1$$
$$\lambda = (2,1) \rightarrow V, \dim = 2$$
Therefore
$$k[S_3] = V_0 \oplus V_1 \oplus V^{\oplus 2}$$
as desired.
If this is new to you, then there are a lot of details to check here, all of which can be found in Fulton Harris. You should know that this story works for any $n$ and we have in general that
$$K[S_n] = \bigoplus_{\lambda \vdash n} V_\lambda^{\oplus \dim V_\lambda}$$
where $V_\lambda$ is the irreducible representation of $S_n$ corresponding to the partition $\lambda$ of $n$.
First of all, the Peter-Weyl Theorem is about unitary representations, it will be of no use if you want to study representations on topological vector spaces which are not isomorphic to Hilbert spaces (or pre-Hilbert spaces). Thus, formally speaking, the answer to your question (at the end of the 1st paragraph) is negative. Nevertheless, one has:
Theorem. Let $V$ be a Hausdorff locally convex quasicomplete topological vector space, $G$ a compact (and Hausdorff) topological group and $\rho: G\to Aut(V)$ a continuous irreducible representation (meaning that $V$ contains no proper closed invariant subspaces). Then $V$ is finite-dimensional.
You can find a proof in
R.A.Johnson, Representations of compact groups on topological vectors spaces: some remarks, Proceedings of AMS, Vol. 61, 1976.
The point of considering this class of topological vector spaces (which includes, for instance, all Banach spaces) is that one has a satisfactory theory of integration for maps to such spaces. As for more general vector spaces, I have no idea, I suppose that the finite-dimensionality claim is simply false. For instance, if you drop the assumption that $V$ is Hausdorff (which one usually assumes) and take a vector space with trivial topology, you will have irreducible representations of any group on such a vector space.
One last thing: In the context of Hilbert spaces, there is a very short and direct (avoiding PWT) proof of finite dimensionality, given in this mathoverflow post.
Best Answer
Let $A=\mathbb{C}[x]$ and consider the regular representation of $A$ acting on itself. Every nonzero element generates a subrepresentation isomorphic to $A$, so this definitely does not have any irreducible subrepresentations.