Let $X, Y$ be Hilbert spaces and $A\in L(X,Y)$ be a bounded linear map. Let $\{ R_t\}_t$ a family of functions $Y \rightarrow X$ and $\gamma: \mathbb{R}_+\times Y\rightarrow \mathbb{R}_+$ with the property $$\sup\{||A^+g-R_{\gamma(\epsilon,z)}z||, g \in \text{domain}(A^+), ||g-z||\leq \epsilon\}=:h(\epsilon) \rightarrow 0$$
where $A^+$ is the Moore-Penrose pseudoinverse.
Is it then true that $A^+$ is continuous?
I would say yes, as if we take $g_n\rightarrow g$ in $\text{domain}(A^+)$, one can trivially write, for $\epsilon_n := ||g_n-g||$ $$||A^+g-A^+g_n||\leq ||A^+g-R_{\gamma(\epsilon_n,g_n)}g_n|| +||R_{\gamma(\epsilon_n,g_n)}g_n-A^+g_n||\\\leq h(\epsilon_n)+h(0)=h(\epsilon_n)\rightarrow0$$
Is this reasoning correct/is my claim correct?
Your proof is correct, however the definition of the regularizing family is different than usually. I found the following definition in the book by Rieder [Keine Probleme mit inversen Problemen (in German)] Def 3.1.1: $$ \forall g\in D(A^\dagger): \ \sup\{ \|A^\dagger g - R_{(\cdots)}(z): \ \|z-g\|_Y\le \epsilon\} \to 0 \quad \text{ for } \epsilon\to 0. $$ You wrote $$\ \sup\{ \|A^\dagger g - R_{(\cdots)}(z): \ g\in D(A^\dagger),\ \|z-g\|_Y\le \epsilon\} \to 0 \quad \text{ for } \epsilon\to 0, $$ which is much stronger, as it implies continuity of $A^\dagger$.