No, but you can say it is *lower semicontiuous*, even wrto the initial time. 

Precisely, given a Banach space $E$, an open set $\Omega\subset \mathbb{R}\times E$ and  $f:\Omega\subset E\times \mathbb{R}\rightarrow E$ in the Cauchy-Lipshitz-Picard hypoteses, for any $(t_0,x_0)\in \Omega$
the Cauchy problem $$u(t_0)=x_0$$ for the ODE  $$\dot u =f( t, u(t))$$ 

admits a maximal solution defined in an interval $\big(\tau_*(t_0,x_0), \tau^*(t_0,x_0)\big)\subset\mathbb{R}$, where 
$$\tau _ *:\Omega\to [-\infty,0 ) $$ 
is upper semicontinuous and 
$$\tau ^ *:\Omega\to (0,+\infty]$$ 
is lower semicontinuous. This amount to saying that: the domain of the "general solution" $\xi:\Xi\subset \Omega\times\mathbb{R}\rightarrow E$ defined as $\xi(t_0,x_0,t):=u(t)$ with the solution $u(t)$ of the above Cauchy problem, that is the set
$$\Xi:=\{ (s,x,t)\in \Omega\times\mathbb{R} \\ :\\ \tau _ * (s,x) < t < \tau ^ *(s,x) \}$$
(that is the zone between the graph of $\tau _ * $ and the graph of $\tau ^ *$), is an open set.