Show the isomorphism $X/\{0\} \cong X$ $X/\{0\}$ is the set of singletons of $X$ so we define a mapping $T(x) = x$. Since $\{0\} \subseteq \ker T$ we have a linear mapping $\hat{T}:X/\{0\}\to X$.
But $\{0\}$ is exactly equal to $\ker T$ so $X/\ker T \cong ran T = X$ since the mapping is surjective. 

Does this make sense? 

So yes, your reasoning seems to be correct. Although it took me a while to fully understand what you are doing here. So let's make it rigorous (for example by giving explicitely domains and codomains of each map).
Let $X$ be a group/ring/module/linear space (or anything that has an analogue of the first isomorphism theorem). Define
$$\mbox{id}:X\to X$$
$$\mbox{id}(x)=x$$
The famous identity map. Now this map is a homomorphism. Therefore by using the first isomorphism theorem we obtain an isomorphism
$$\overline{\mbox{id}}:X/\ker(\mbox{id})\to\mbox{id}(X)$$
$$\overline{\mbox{id}}(x\ker(\mbox{id}))=x$$
Since $\ker(\mbox{id})=\{0\}$ and $\mbox{id}(X)=X$ then this gives us an isomorphism
$$\overline{\mbox{id}}:X/\{0\}\to X$$
