Totally path disconnected spaces are $T_1$ 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?
 A: 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.
A: Let X be  a totaly path disconnected space and let $a,b \in X$ be distinct points.
Define $f:[0,1] \to X$ by
$$f(x) = \begin{cases} b, & x \in [0,1) \\ a, & x=1 \end{cases}$$
Notice that this function can't be continous as otherwise $a$ and $b$ are in the same path component.
Thus there exist an open set $U$ such that $f^{-1}(U)$ is not open,now the preimages of $f$ are $\emptyset, [0,1), \{1\},X$ and the only one that is not open is $\{1\}$ and thus $f^{-1}(U) = \{1\}$ so $U$ has to contain $a$ and not $b$.
And thus $X$ is $T_1$
