Let $D \subseteq \mathbb{R}\times\mathbb{R}^n$ be open, and let $f_m$ be a sequence in $C(D, \mathbb{R}^n)$ that converges uniformly on compact subsets of $D$ to $f_0 \in C(D, \mathbb{R}^n)$. Let $(t_m, a_m)$ be a sequence in $D$ that converges to $(t_0, a_0) \in D$. For each $m \in \mathbb{N} \cup \{0\}$, let $x_m$ be a solution of the IVP $$\frac{dx}{dt} = f(t, x), \qquad x(t_m) = a_m.$$ Show that there exists $\delta > 0$ and $N \in \mathbb{N}$ such that for all $m \geq N$, we have $[t_0 - \delta, t_0 + \delta] \subseteq I(x_m)$, where $I(x_m)$ denotes the maximal interval of existence for $x_m$.
Using the fact that $t_n \to t_0$, I can see that $t_n \in I(x_0)$ for sufficiently large $n$, so at least $I(x_m) \cap I(x_0)$ has non-empty interior. However, I am having trouble showing even something like $t_0 \in I(x_m)$ eventually; at first, it seems that each $I(x_m)$ could 'avoid' the point $t_0$. If I could find some $t \in I(x_m)$ for all $m \geq N$, I think I can use the form $$ x_m(t) = a_m + \int_{t_m}^t f_m(s, x_m(s)) \:ds $$ to show that $x_m(t) \to x_0(t)$, but this feels circular at this point.