Let $B$ be a Banach space (in particular, $B$ is a function space equipped with the supremum norm). The space $C^0 ([-T,T],B)$ is the set of continuous functions on $[-T,T]$, valued in $B$.
Question 1: Am I right in thinking that the "boundedness theorem" is true in this case? That is, if I have $f \in C^0([-T,T],B)$, then there is a $K$ such that
$$\| f(t) \|_{B} \leq K \quad \forall t\in[-T,T]. $$
My idea here is that since $\| \cdot \|:B\to \mathbb{R}_{\geq 0}$ is continuous, the composition $\|\cdot\| \circ f: [-T,T] \to \mathbb{R}_{\geq 0}$ is also continuous. With $[-T,T]$ compact, the boundedness theorem applies. Is this correct?
Question 2: Is the space $C^0([-T,T],B)$ a Banach space?
If I am correct in question 1, I think this follows if I define the infimum of all $K$ to be the norm of $f$. I think $C^0([-T,T],B)$ will be a Banach space under this norm.
Question 1: yes, $f$ is bounded. Your reasoning is OK. Alternatively you could argue like this: if $f:X\rightarrow Y$ is a continous map between topological spaces, then the image $f(K)$ of a compact set $K\subseteq X$ is compact. Moreover in a metric space compact sets are bounded.
Question 2: yes, you are right $\|f\|:=\max (|f(t)| : t\in[-T,T])$ and $C^0([-T,T],B)$ becomes a Banach space under this norm. In fact completeness is equivalent to the completeness of $B$.