No, there is an $\pi_1$ issue. Let $G$ be an acyclic group, then the classifying space BG is a space whose homology vanishes, but it is not contractible and if G is infinite than it is not homotopy equivalent to a finite complex. An example of such an infinite acyclic group is the group G of all bijections of an infinite set.

Taking the line with two origin's example further, as Thomas Rot points out the two different origins in that space are necessarily in any psuedo-metric open ball (centered at one of them say). They cannot be separated by open sets so the topology generated by the pseudo metric is not $T_0$. In particular the pseudometric topology is different from the topology as a non-Hausdorff manifold.

As far as I know it is an open question whether Turaev-Viro tqfts can distinguish L(7,1) and L(7,2). I was just pointing out that the category that was expected to do this seems not to work.

Have you seen this: mathoverflow.net/questions/239851/… ?

@DavidRoberts That is a fair point. I edited to use $Cat_{(\infty,0)}$.

There is a variant of the notion of plot where you only require $\tilde{p}: M \times [0,1] \to \mathbb{R}$ to be continuous (but if you fix $t$, $p(t)$ is still smooth with compact support in the $x$-direction). I think what your argument really shows is that any $\tau_2$-continuous map $p: [0,1] \to C^\infty_c(\mathbb{R})$ has to be a plot of this kind.

@FanZheng This is regarding your potential counter-example to $\tau_2 \subseteq \tau_4$ two comments above. Your $p(t,x)$ does not define a plot. It is true that for each $t$ $p(t)$ is a compactly supported function, but the requirement to be a plot is that there is a single compact $K$ so that all the functions in the path are supported in $K$. As $t \to 0$ your $p(t)$ is supported on larger and larger subsets which contain at least $[-\frac{1}{t}, + \frac{1}{t}]$. Hence your $p(x,t)$ is not a plot and there is no counter example.