This is a variant of the question Compact Embedding of $W^{1,2}(0,T;ℝ^d)$ in $C(0,T;ℝ^d)$ where we had $X=\mathbb{R}^d$. Let now $X$ be some Banach space.
Question: Is $H^1(0,T;X)$ compactly embedded in $C([0,T];X)$?
We define $H^1(0,T;X)=\{u : (0,T) \to X ~|~ u \in L^2(0,T;X), u' \in L^2(0,T;X)\}$.
A proof for the continuous embedding can be found in a lot of books, e.g. Roubicek's book. Also, it is standard that $H^1(0,T,\mathbb{R})$ is compactly embedded in $C([0,T];\mathbb{R})$, e.g. Brezi's book. The main difficulty is that the classical Arzela-Ascoli theorem is only valid for handling something like $C(S;\mathbb{K}^d)$ and not for $C(S;Y)$ where $Y$ is some infinite-dimensional Banach space. But for this scenario there is the following version, which can be found in the book of Boyer.
Arzela-Ascoli: Let $E$ be a compact metric space and $F$ a metric space. Let $\mathcal{F}$ be a subset of $C(E,F)$. If $\{f(x) : f \in \mathcal{F} \}$ is relatively compact in $F$ for any $x\in E$ and $\mathcal{F}$ is equicontinuous, then $\mathcal{F}$ is relatively compact in $C(E,F)$.
I am translating the proof of Brezis to our setting.
Proof: Let $u \in B:=\{v \in H^1(0,T;X) : \|v\|_{L^2(0,T;X)}+\|v'\|_{L^2(0,T;X)} \leq 1\}$. In particular $u \in C([0,T];X)$. Then we have for all $t,s \in [0,T]$ $$\|u(t)-u(s)\|_X \leq \int_s^t \|u'(\tau)\|_X \, \text{d}\tau \leq \|u'\|_{L^2(0,T;X)} |t-s|^{1/2} \leq |t-s|^{1/2}$$ i.e. $u$ is equicontinuous. Further if we can show that $\{ u(t): u \in B\}$ is relatively compact in $X$ for any $t\in [0,T]$, then we would be finished.
Can someone help me with this last step written in bold font? I am not sure if this result is even correct. I think I have seen it in a article with $X=L^2$ or $X=H^1$, I don't remember clearly.
Before there was a picture of another Arzela-Ascoli theorem version which was written in an unclear/erroneous way. I've updated it with the version in Boyer's book which coincides with the notes provided in the comments.
Having infinite-dimensional $X$ is problematic, because even the space of constant $X$-valued functions is not compactly embedded in $C([0, 1];X)$. Indeed, the image of its unit ball under the embedding into $C([0,1];X)$ is an isometric copy of the unit ball of $X$; not a totally bounded set.
The result you are looking for is the Aubin–Lions(-Simon) lemma where an additional compactness condition is imposed on the values taken by the functions in our space.