Is it possible to have a tubular neighbourhood of a $C^0$ submanifold? For instance, consider the embedding of a unit square in $\Bbb R^2$,
\begin{eqnarray} f : S &\to& \Bbb R^2\\ t &\mapsto& \begin{cases} (\frac 1 2, \frac 12 - t) & t \in [0,1]\\ (\frac 1 2 - (t - 1), - \frac{1}2) & t \in [1,2]\\ (-\frac 1 2, - \frac{1}2 + (t - 2)) & t \in [2,3]\\ (-\frac 1 2 + (t - 3), \frac{1}2) & t \in [3,4]\\ \end{cases} \end{eqnarray}
Does there exist a neighbourhood $N \approx S \times \mathbb R$ with
\begin{eqnarray} g : S \times \Bbb R \to \Bbb R^2 \end{eqnarray} an injective function such that there is a projection $g(t, 0)$ mapping to the square itself?
The key condition is "local flatness". A topological $k$-dimensional submanifold $K\subset R^n$ (more generally, a submanifold of an $n$-dimensional manifold $M$) is called locally flat if for every $x\in K$ there exists a neighborhood $U$ of $x$ in $R^n$ and a homeomorphism $\phi: U\to V\subset R^n$ such that $\phi(K\cap U)=V\cap R^k$. It is a nontrivial fact (Schoenflies theorem) that every (embedded) topological submanifold in $R^2$ is locally flat. On the other hand, $R^3$ contains non-locally flat surfaces (see here and circles (see here for instance).
On the other hand, Morton Brown proved that every locally flat topological submanifold in a topological manifold, admits a tubular neighborhood:
Morton Brown, "Locally flat imbeddings of topological manifolds", Annals of Mathematics, Vol. 75 (1962), p. 331-341.
Here an (open) tubular neighborhood of a submanifold $S\subset M$ is an open subset $U\subset M$ containing $S$ such that there exists a locally trivial topological disk bundle $U\to S$ such that $S\subset U$ is the image of a section of this bundle.
The proof is easier (but still nontrivial), of course, in the case of topological curves in $R^2$ (which, as I said, are all locally flat).
Lastly, in the case of the square (as in the case of any convex Jordan curve $C\subset R^2$), the proof of the existence of a tubular neighborhood is elementary, in fact, there exists a homeomorphism $R^2\to R^2$ sending such a curve to the unit circle.