Andrew takes a square sheet of paper $ABCD$ of side length $1$ and folds a kite shape. To do this, he takes the corners at $B$ and $D$ and folds the paper such that both corners now rest at a point $E$ on $AC$. This fold results in two creases $CF$ and $CG$, respectively, where $F$ lies on $AB$ and $G$ lies on $AD$. Compute the length of $FG$.
1. **Understanding the problem**: We start with a square sheet of paper \(ABCD\) with side length 1. Andrew folds the paper such that corners \(B\) and \(D\) meet at a point \(E\) on the diagonal \(AC\). This creates two creases \(CF\) and \(CG\), where \(F\) lies on \(AB\) and \(G\) lies on \(AD\). We need to find the length of \(FG\).

2. **Analyzing the fold**: Since \(B\) and \(D\) are folded to meet at \(E\), \(E\) must lie on the diagonal \(AC\). The creases \(CF\) and \(CG\) bisect the angles \(\angle ACB\) and \(\angle ACD\) respectively, making \(\angle ECF = \angle ECG = 22.5^\circ\) (since \(\angle ACB = \angle ACD = 45^\circ\)).

3. **Using trigonometry**: Let \(EF = EG = x\). Since \(\angle ECF = 22.5^\circ\), we can use the tangent function:
   \[
   \tan(22.5^\circ) = \frac{x}{1} = x
   \]
   Using the tangent half-angle formula, we know:
   \[
   \tan(22.5^\circ) = \sqrt{2} - 1
   \]
   Therefore:
   \[
   x = \sqrt{2} - 1
   \]

4. **Finding \(FG\)**: Since \(F\) and \(G\) are reflections of each other across \(E\), the length \(FG\) is twice the length of \(EF\):
   \[
   FG = 2x = 2(\sqrt{2} - 1) = 2\sqrt{2} - 2
   \]

5. **Verification with another approach**: Let \(x\) be the length of \(FG\). Notice that side \(AB\) has been reflected onto line segment \(AE\) after the folds, hence \(AE = 1\). Furthermore, because \(\triangle FGC\) is a \(45^\circ-45^\circ-90^\circ\) triangle, we get that \(EC = EF = \frac{FG}{2} = \frac{x}{2}\). Since \(AE + EC\) is the diagonal of the square, we get the equation:
   \[
   1 + \frac{x}{2} = \sqrt{2}
   \]
   Solving for \(x\):
   \[
   1 + \frac{x}{2} = \sqrt{2} \implies \frac{x}{2} = \sqrt{2} - 1 \implies x = 2(\sqrt{2} - 1) = 2\sqrt{2} - 2
   \]

Conclusion:
\[
FG = 2\sqrt{2} - 2
\]

The final answer is \(\boxed{2\sqrt{2} - 2}\)