No, but you can say it is lower semicontiuous, even wrto the initial time (that is, the optimistic situation: perturbing a little the initial data the existence is ensured almost up to $T$ , and could even be much greater) .
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.