Step 1: Know the target shape.
A see-saw shape comes from a trigonal bipyramidal electron arrangement (steric number $5$) that has exactly one lone pair sitting in an equatorial spot. So we hunt for the species with steric number $5$ and one lone pair.
Step 2: Set up the counting rule.
Steric number $= \frac{1}{2}(\text{central atom valence electrons} + \text{number of single-bonded atoms} - \text{cation charge} + \text{anion charge})$. Lone pairs $=$ steric number minus the number of bonded atoms.
Step 3: Test $\text{SF}_4$.
Sulphur has $6$ valence electrons and bonds to $4$ fluorines. $\text{SN} = \frac{1}{2}(6+4) = 5$, with $4$ bond pairs and $1$ lone pair. That is trigonal bipyramidal with one lone pair, giving a see-saw shape. This already fits, but we confirm the others to be sure.
Step 4: Test $\text{XeF}_4$.
Xenon has $8$ valence electrons and bonds to $4$ fluorines. $\text{SN} = \frac{1}{2}(8+4) = 6$, with $4$ bond pairs and $2$ lone pairs, giving a square planar shape, not see-saw.
Step 5: Test $\text{CCl}_4$ and $\text{BF}_4^-$.
For $\text{CCl}_4$, $\text{SN} = \frac{1}{2}(4+4) = 4$ with no lone pair, so it is tetrahedral. For $\text{BF}_4^-$, $\text{SN} = \frac{1}{2}(3+4+1) = 4$ with no lone pair, also tetrahedral. Neither is see-saw.
Step 6: Conclude.
Only $\text{SF}_4$ has the trigonal bipyramidal core with a single equatorial lone pair, so it alone shows the see-saw shape, which is option (A).
\[ \boxed{\text{SF}_4 \text{ is see-saw}} \]