Consider $X_0\leftarrow X_1\leftarrow \dots$ be a sequential diagram in a category $C$ and suppose it has a limit. I want to show that if $(n_k)$ is a sequence of increasing natural integers, then the sequential diagram $X_{n_0}\leftarrow X_{n_1}\leftarrow \dots$ also has a limit and the natural morphism lim$_nX_n\to $lim$_kX_{n_k}$ is an isomorphism.
I have two questions, first what does it mean for this diagram to have a limit ? The definition I have which is the most fitting is that if $F:Nat^{op}\to D$ is a functor with a limit, then we call it the inverse limit of the tower $F(0)\leftarrow F(1)\leftarrow \dots$ Here $Nat$ is the category of natural integers, with $Nat(m,n)=\{j_{m,n}\}$ if $m\leq n$ and is empty otherwise. Do I assume that there is such a functor with $F(i)=X_i$ ?
Secondly where does the natural map lim$_nX_n\to $lim$_kX_{n_k}$ come from ? Is it because lim$_nX_n$ will satisfy the universal property of the limit for the second sequential diagram ?
Yes, by writing the tower $X_0\leftarrow X_1 \leftarrow …$ you are actually giving the datum of a functor $\mathsf{Nat}^{op} \rightarrow C$ and a limit of the tower is meant to be a limit of this functor/diagram.
The canonical map $\lim_n X_n \rightarrow \lim_k X_{n_k}$ comes from the fact, that any cone over the diagram $(X_n)_n$ automatically also is a cone over the diagram $(X_{n_k})_k$. In particular the limit cone of the $(X_n)_n$ becomes a cone over the $(X_{n_k})_k$ and thus admits an induced arrow to the limit cone over the latter.
To show that this induced morphism is an isomorphism, it suffices to note that any cone over the $(X_{n_k})_k$ can be extended to a cone over the $(X_n)_n$. Indeed, if $C\Rightarrow (X_{n_k})$ is a cone and $m \neq n_k \forall k$, such that the cone leg $C\rightarrow X_m$ is missing, pick some $k$ such that $n_k > m$ and let the missing leg be the composite $C \rightarrow X_{n_k} \rightarrow X_m$, where the first map is the corresponding leg of the existing cone and the second morphism is the transition map of our diagram. This is welldefined (and clearly necessary to be part of a cone over the whole diagram). In fact, restriction and extension now give us an isomorphism of categories of cones over the diagrams $(X_n)_n$ and $(X_{n_k})_k$. In particular this isomorphism preserves terminal objects, ie. the limit cones over the respective diagrams.
Edit: In the second paragraph I assume the limit of the diagram $(X_{n_k})_k$ to exist, to answer the second question. However the argument in the third paragraph works without this assumption. This is to say: The isomorphism of categories of cones shows that the limit of the diagram $(X_n)_n$ exists if and only if the limit of the diagram $(X_{n_k})_k$ exists and that we can compute the limit cones as the restriction / extension of the limit cones.