Color the points in $S_n$ as follows (see Figure 1):
- if $y \ge 0$, color $(x, y)$ white if $x + y - n$ is even and black if $x + y - n$ is odd;
- if $y < 0$, color $(x, y)$ white if $x + y - n$ is odd and black if $x + y - n$ is even.

Figure 1: Coloring of $S_3$

Consider a path $(x_1, y_1), (x_2, y_2), \dots, (x_\ell, y_\ell)$ in $S_n$. A pair of successive points $(x_{i-1}, y_{i-1})$ and $(x_i, y_i)$ in the path is called a pair of successive black points if both points in the pair are colored black.

Suppose now that the points of $S_n$ are partitioned into $m$ paths and the total number of successive pairs of black points in all paths is $k$. By breaking the paths at each pair of successive black points,

we obtain $k+m$ paths in each of which the number of black points exceeds the number of white points by at most one. Therefore, the total number of black points in $S_n$ cannot exceed the number of white points by more than $k+m$. On the other hand, the total number of black points in $S_n$ exceeds the total number of white points by exactly $2n$ (there is exactly one more black point in each row of $S_n$). Therefore,
$$2n \leq k + m.$$
There are exactly $n$ adjacent black points in $S_n$ (call two points in $S_n$ adjacent if their distance is 1), namely the pairs
$(x, 0)$ and $(x, -1)$,
for $x = -n+1, -n+3, \dots, n-3, n-1$. Therefore $k \leq n$ (the number of successive pairs of black points in the paths in the partition of $S_n$ cannot exceed the total number of adjacent pairs of black points in $S_n$) and we have $2n \leq k+m \leq n+m$, yielding
$$n \leq m.$$