You are mixing two things here:
1) In general there is only a unique homomorphism not an isomorphism.
2) The inverse limit is still unique up to unique isomorphism.
The reason that you still have 2) is that if you had inverse limits $X$ and $Y$ then you had the universal property for both of them. So you have unique homomorphisms $X\to Y$ and $Y\to X$. You can show that their compositions are the identities (by invoking the universal property for a third time.)
Edit To clarify regarding the comments: You have the universal propery for an object $X$. Then you can test it for any objects $Y$ which comes with the according $\pi_i$. You get a homomorphism $f:Y\to X$. Note that $Y$ can be a fairly arbitrary object. Only if $Y$ also has the universal property you get a map in the other direction which then turns out to be an inverse.
Edit2: Your example is pretty good so I'll work with it. So let $X_n=[-\frac1n,\frac1n]$ with the obvious inclusions as maps between them. Which candidates for the inverse limit do we have? First note that $X$ comes with maps (i.e. inclusions) to all of the $X_n$, so $X$ is a subset of every $X_n$ and therefore a subset of the intersection. In our example $X$ can only be $\{0\}$ or $\emptyset$, since those are the only subsets of the intersection. Note that the class of all candidates at this point is the class of all "$Y$" at which we have to test in the next step.
Now we turn to the universal property: for any test-set $Y$ which is a subset of the intersection we want to have a morphism (inclusion) $Y\subset X$. Thus $X$ must be contained in the intersection (step 1) and also contain all other sets with this property (step 2), which leaves us with $X=\{0\}$.
This is somewhat the whole idea of what is going on there. In general $X$ has to be "small enough" to come with maps into all the $X_n$, and then it must be the "biggest" among all objects with this property.