Yes, it's not hard to show that $f$ is Lipschitz of constant 
$$k:=\sup_t\lim\sup_{s\to t}\frac{|f(s)-f(t)|}{|t-s|}\\ .$$ 

Indeed, let's take any $l > k$.  By assumption, any $t\in I:=[0,1]$ has a nbd $J_t:=]t',t''[\cap I$ such that for any $s\in J_t$ one has $|f(s)-f(t)|\le l|t-s|$. Extract a finite covering of $I$ by these nbd's, say $J_{t_0}\dots J_{t_{n }}$  with $0:=t_0 < t_1 < t_2 <\dots < t_n:= 1$, and assume it is minimal. By minimality, it follows $J_{t_i}\cap J_{t_{i+1}}\neq\emptyset$ for all $0\le i < n$. As a consequence,  there is a sequence $0:=s_0 \le s_1 \le \dots \le s_{2n}:=1$ such that $s_{2i}=t_i$ and $s_{2i+1}\in J_{t_i}\cap J_{t_{i+1}}$, Therefore, since any two consecutive $s_j$'s are (in some order) a $t_i$ and an element of $J_{t_i}$, we have
$$|f(1)-f(0)|\le \sum_{i=0}^{2n-1}|f(s_{i+1}) - f(s_i)|\le l\sum_{ i = 0 }^{2n-1} (s_{i+1} - s_i)=l\\ ,  $$
and in fact $|f(1)-f(0)|\le k$ because the above inequality holds for any number $l > k$. 
This also implies, by rescaling and localizing, 
$$|f(b)-f(a)|\le k |b-a|$$
for all $a$ and $b$ in $I$.