The result you quote is part of the main theorem of the book.   I can look it up later today, but it's one part of the proof that every regular Dirichlet form has an associated Markov process.

The problem with your argument, in two words, is null sets.  $T_t$ is an operator on $L^2$, and an element of $L^2$ isn't a function, it's an equivalence class mod a.e. equality.  You would like to say "For each $x \in X$, consider the linear functional on $C_0(X)$ defined by $f \mapsto T_t f(x)$, and then use Riesz-Markov-Kakutani to find the measure associated to this functional and call it $P_t(x, \cdot)$."  But $T_t f$ is only given as an element of $L^2$, and so $T_t f(x)$ isn't well defined; it depends on which representative of $T_t f$ you choose.

I think FOT has some counterexamples in the case that the form is not regular.  

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Basically the issue is that your state space $X$ may have a set $A$ which is measure zero but positive capacity.  Then $X \setminus A$ looks just the same as a measure space, so the semigroup on $L^2$ still makes sense, but the transition function now wants to send the process into $A$ but can't tell you what to do after that.  I can try to dig up some old notes I have if you don't find it in FOT.
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