This is false even in the case you describe as it would imply Cantor-Schroeder-Bernstein for fields. That said, we can the standard counter example for the claim of CSB for fields as a counter example for your claim. Let $F_i = \mathbb{C}(X)$ and $K_i = \mathbb{C}$ for all $i.$ Then $K_i$ injects into $F_i$ and $F_i$ injects into $K_i$ for all $i,$ but the unions $F$ and $K$ are not isomorphic, as $F$ is not algebraically closed.

This is false even in the case you describe as it would imply Cantor-Schroeder-Bernstein for fields. That said, we can take the standard counter example for the claim of CSB for fields as a counter example for your claim. Let $F_i = \mathbb{C}(X)$ and $K_i = \mathbb{C}$ for all $i.$ Then $K_i$ injects into $F_i$ and $F_i$ injects into $K_i$ for all $i,$ but the unions $F$ and $K$ are not isomorphic, as $F$ is not algebraically closed.

This is false even in the case you describe as it would imply Cantor-Schroeder-Bernstein for fields. One can take the standard counter example for proof. Let $F_i = \mathbb{C}(X)$ and $K_i = \mathbb{C}$ for all $i.$ Then $K_i$ injects into $F_i$ and $F_i$ injects into $K_i,$ but $F$ and $K$ are not isomorphic, as $F$ is not algebraically closed.

This is false even in the case you describe as it would imply Cantor-Schroeder-Bernstein for fields. That said, we can the standard counter example for the claim of CSB for fields as a counter example for your claim. Let $F_i = \mathbb{C}(X)$ and $K_i = \mathbb{C}$ for all $i.$ Then $K_i$ injects into $F_i$ and $F_i$ injects into $K_i$ for all $i,$ but the unions $F$ and $K$ are not isomorphic, as $F$ is not algebraically closed.