A totally path disconnected space is a space $X$ whose path-connected components are all singletons. In other words, all continuous functions $[0,1]\to X$ must be constant.
Every totally disconnected space is totally path disconnected (since path components are contained in connected components). And totally disconnected spaces are $T_1$ (since connected components are always closed).
But the following more general result is also true:
Proposition: Every totally path disconnected space is $T_1$.
This will allow to update a result in pi-base. Can anyone provide a proof?
Best Answer
If $X$ is totally path disconnected, then obviously every subspace of $X$ is totally path disconnected. In particular, every two-element subspace is totally path disconnected, hence discrete. Thus, $X$ is T1.
Additional note as requested:
if $X = \{0, 1\}$ is totally path disconnected, then it is discrete:
Assume not, then w.l.o.g. $\{1\}$ is not open. Hence $f: [0,1] \rightarrow X, f(x) = \begin{cases} 0, & x < \frac{1}{2} \\ 1, & x \ge \frac{1}{2} \end{cases}$
is continuous, but not constant.
Obviously, a space is T1, iff every two-element subspace is T1, iff every two-element subspace is discrete.