Let $f:[0,T] \to \mathbb{R}^n$ be a continuous function, I have the intuition that the following set is compact: $$ A:=\bigcup_{t\in [0,T]}\overline{B}(f(t),1) $$ with $\overline{B}(f(t),1)$ is the closed ball of center $f(t)$ and raduis $1$. But, I do not know how to prove it.
My idea: is that since we are in finite dimension, then, it sufficies to prove that is closed and bounded. It is obvious that it is closed (as the union of closed sets). But I am struggling with the compactness. I think of the following argument $$ A\subset \overline{B}(0, 2M) $$ where $M:=\sup_{t\in[0,T]}\|f(t)\|$.
Your question can be generalized as follows:
Let $C \subset \mathbb R^n$ be compact. Show that $$C' = \bigcup_{x \in C} \overline B(x,1)$$ is compact.
In fact $C'$ is bounded. An upper bound is $M + 1$ with $M = \sup_{x \in C} \lVert x \rVert$. Taking $2M$ is not a good idea because $M < 1$ is possible.
It remains to prove that $C'$ is closed. Note that unions of closed sets are in general not closed, thus a better argument is needed.
Let $(x'_n)$ be a sequence in $C'$ converging to some $\xi' \in \mathbb R^n$. We have to show that $\xi' \in C'$. Choose $x_n \in C$ such that $x'_n \in \overline B(x_n,1)$. Since $C$ is compact, $(x_n)$ has a subsequence $(x_{n_k})$ converging to some $\xi \in C$. Clearly $x'_{n_k} \to \xi'$. For each $m \in \mathbb N$ there exists $k_0(m)$ such that $\lVert \xi -x_{n_k} \rVert < 1/m$ and $\lVert \xi' -x'_{n_k} \rVert < 1/m$ for $k \ge k_0(m)$. For each $m$ we get with $k \ge k_0(m)$ $$\lVert \xi' - \xi \rVert \le \lVert \xi' -x'_{n_k} \rVert + \lVert x'_{n_k} -x_{n_k} \rVert + \lVert x_{n_k} - \xi \rVert < 1/m + 1 + 1/m = 1 + 2/m.$$ This is possible only when $\lVert \xi' - \xi \rVert \le 1$, thus $\xi' \in \overline B(\xi,1) \subset C'$.