Step 1: Understand the question.
We must balance $x\text{H}_2\text{O}_2 + \text{ClO}_4^- \rightarrow x\text{O}_2 + \text{ClO}_2^- + 2\text{H}_2\text{O}$ and find the value of $x$.
Step 2: Write the oxidation half.
In $\text{H}_2\text{O}_2$ oxygen is $-1$, and in $\text{O}_2$ it is 0, so oxygen is oxidised and loses electrons.
\[ \text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2e^- \]
Step 3: Write the reduction half.
Chlorine drops from $+7$ in $\text{ClO}_4^-$ to $+3$ in $\text{ClO}_2^-$, a gain of 4 electrons.
\[ \text{ClO}_4^- + 4\text{H}^+ + 4e^- \rightarrow \text{ClO}_2^- + 2\text{H}_2\text{O} \]
Step 4: Match the electrons.
The oxidation half gives 2 electrons but the reduction half needs 4. So multiply the oxidation half by 2.
\[ 2\text{H}_2\text{O}_2 \rightarrow 2\text{O}_2 + 4\text{H}^+ + 4e^- \]
Step 5: Add the halves.
The 4 electrons and 4 $\text{H}^+$ cancel on both sides.
\[ 2\text{H}_2\text{O}_2 + \text{ClO}_4^- \rightarrow 2\text{O}_2 + \text{ClO}_2^- + 2\text{H}_2\text{O} \]
Step 6: Read x.
The coefficient in front of $\text{H}_2\text{O}_2$ and $\text{O}_2$ is 2, so $x = 2$. This is option 4.
\[ \boxed{x = 2} \]