I am trying to solve the following problem:
I unfortunately don't even know where to start. Suppose there is some model $\mathcal{A}$ such that $\mathcal{A}\vDash T$, and there is some elementary extension $\mathcal{A}\prec\mathcal{B}$ with $\mathcal{B}\vDash T'$.
For sake of contradiction, assume there is some $\mathcal{M}$ such that $\mathcal{M}\vDash T$, but for all $\mathcal{N}$ with $\mathcal{M}\prec\mathcal{N}$, we do not have $\mathcal{N}\vDash T'$. Let $\mathcal{N}$ be any elementary extension of $\mathcal{M}$ (exists by upward L-S), then there is some $\sigma\in T'$ such that $\mathcal{N}\not\vDash \sigma$.
After this I don't know how to proceed. Since $T$ is complete, we know $\mathcal{A}$ and $\mathcal{M}$ agree on all $\mathcal{L}$-sentences (but since we know nothing of their cardinality, we can't conclude they are isomorphic, right?). So the two models of $T$ are connected, and we can elementarily-extend from either of those into $\mathcal{B}$ or $\mathcal{N}$.
But I don't see how to generate a contradiction using $\sigma$, as it is an $\mathcal{L}'$-sentence so I don't get how to use $\mathcal{A}$ or $\mathcal{M}$ to talk about it. Is there a compactness argument here somewhere? Any guidance would be much appreciated.
Let $\mathcal M\models T$ and use compactness to show that the elementary diagram of $\mathcal M$ is consistent with $T'.$
A finite subtheory will be equivalent to a sentence of the form $\varphi(\vec c)\land \psi$ where $\psi\in T'$ and $\varphi(\vec c)$ is some $\mathcal L\cup\{\vec c\}$-sentence that is true in $\mathcal M$ when $\vec c$ are interpreted as the corresponding elements of $\mathcal M.$ Then, since $T$ is complete and $\mathcal B\models T,$ there will be some $\vec d\in \mathcal B$ such that $\mathcal B\models \varphi(\vec d)\land \psi.$
So then $\mathcal B$ satisfies $\varphi(\vec c)\land \psi$ when you assign the constant symbols $\vec c$ to $\vec d$.