We have an infinite sequence of real numbers $(a_n)_{n \geq 0}$ satisfying the recursive relation:

$$a_n = \left|a_{n+1} - a_{n+2}\right|$$

for all $n \geq 0$, where $a_0$ and $a_1$ are two different positive real numbers. Our goal is to determine whether this sequence can be bounded.

### Step 1: Analyze the recursive relation
The given relation

$$a_n = \left|a_{n+1} - a_{n+2}\right|$$

implies a functional dependency between three consecutive terms of the sequence. Importantly, since $a_n$ is defined as the absolute difference between $a_{n+1}$ and $a_{n+2}$, the sequence heavily depends on the initial values $a_0$ and $a_1$.

### Step 2: Investigate the effect of positive differences
Since $a_0$ and $a_1$ are given as different positive real numbers, we can start by observing:

- If $a_0 = \left|a_1 - a_2\right|$ and $a_0 > 0$, then either $a_1 > a_2$ or $a_2 > a_1$.
- Similarly, $a_1 = \left|a_2 - a_3\right|$ and $a_1 > 0$ implies that $a_2 \neq a_3$.

### Step 3: Recursive impact of differing terms
Because $a_0$ and $a_1$ are distinct, as the sequence progresses:

- If $a_{n+1} = a_{n+2}$, then $a_n = 0$, which cannot sustain for all $n$ since the sequence starts with non-zero values.
- Given the recursive nature, each $a_n$ serves as a source for subsequent terms, making it improbable for all terms to converge to a single bound due to iterative differences.

### Step 4: Sequence dynamics based on initial conditions
Given that $a_0$ and $a_1$ are positive and different, the sequence infinitely alternates in such a manner that it is impossible for it to stabilize or converge to a bound. The discrepancy between consecutive elements will propagate due to the nature of the absolute difference, thereby causing an unbounded progression along the sequence.

#### Conclusion
Based on the conditions and recursive relation, the sequence $(a_n)$ cannot be bounded, implying that it will diverge or fluctuate indefinitely without settling within any finite bounds.

Therefore, the sequence is not bounded:

$$\boxed{\text{No}}$$