Suppose $u\in L^2(0,T; V)$ and $u_t\in L^2(0,T;V')$, where $V\subset H\subset V'$ is a Gelfand (or Hilbert) triple (the embeddings are dense and continuous and the spaces are all Hilbert). For instance take $H^1_0\subset L^2\subset H^{-1}$. Then it is well-known that $u\in C([0,T];H)$ (see for instance Theorem 6.41 in Hunter's notes: https://www.math.ucdavis.edu/~hunter/pdes/ch6A.pdf)
What goes wrong if we take $V=H$ above (so that instead $V\subset V\subset V'$), and conclude that in fact $u\in C([0,T];V)$? I believe this is incorrect and that the issue stems from how we identify dual spaces, say $H\equiv H'$ and not $V\equiv V'$, but it's unclear exactly how things go wrong.
Because in the triplet $V \subset V \subset V'$ you no longer have $u_t \in L^2(0,T; V')$.
In fact, denote by $I$ the embedding from $V$ to $H$. Then, the statement $u_t \in L^2(0,T;V')$ really means $(I^* u)_t \in L^2(0,T;V')$. By changing the middle space in the triplet, you change $I$ and hence, $u_t \in L^2(0,T;V')$ is no longer valid.