Is the following true? I cannot see a counterexample and it seems very intuitively clear, at least in the embedded case.


Claim:
Consider the set $S$ of closed immersed Riemann surfaces $\Sigma \subset (X,g)$ (I am particularly interested in spheres), with the magnitude of the mean curvature bounded from above by some $C>0$, where $X$ is also closed. Let $Vol(\Sigma)$ denote the area, and $Diam(\Sigma)$ the diameter. Then for every $\epsilon >0$ there is a $\delta>0$ s.t. if
$$Vol (\Sigma) < \delta, \text{ then } Diam (\Sigma)< \epsilon $$ where $\Sigma \in S$.


I really just want to say here that diameter of $\Sigma$ can be assured to be arbitrarily small by requiring that its volume is small.


edit 1: For more clarity let me add that $\epsilon, \delta$ above are assumed to depend on $\Sigma, X, g$.


edit 2: $C$ should be upper bound for magnitude of the mean curvature.



Diameter of immersed surfaces with bounded from above mean curvature
Assuming I am interpreting your question correctly, this is true when the ambient space is Euclidean and the submanifold is closed (i.e. compact and without boundary). To see this one may invoke a result of Topping that bounds the (intrinsic) diameter of a closed, immersed submanifold of dimension $m$ by the $L^{m-1}$ norm of mean curvature over the immersed submanifold, e.g. for a surface take the $L^1$ norm (this scales correctly). 


For a general ambient Riemannian manifold, by isometricly embedding the ambient metric in Euclidean space, one should be able to get a related bound as long as the embedding (of the ambient space) has bounded mean curvature.


See this [answer](https://mathoverflow.net/questions/75953/estimates-for-the-diameter-of-a-nice-surface/76014#76014) to a related question for the reference. 


