What is wrong with how I am approaching this conditional expectation problem 
An individual experiences a loss due to property damage and a loss due
  to bodily injury. Losses are independent and uniformly distributed on the
  interval $[0,3]$. Calculate the expected loss due to bodily injury, given that at
  least one of the losses is less than $1$.

This is my reasoning:
Let $X$ be the property damage loss and $Y$ be the bodily injury loss.
The probability that at least one of the losses is less than $1$ is equal to the sum of the following three probabilities:
$$P(X < 1 , Y > 1) + P(X > 1 , Y < 1) + P(X < 1 , Y < 1)$$
From independence
$${1\over 3} \cdot {2\over 3} + {2\over 3} \cdot {1\over 3} + {1\over 3} \cdot {1\over 3} = {5 \over 9}$$
The probability distribution of $Y$ given that at least one of the losses is less than $1$ is the three cases divided by the sum of all the probabilities that at least one of the losses is less than $1$.
$${P(X < 1 , Y > 1)\over {5 \over 9}} ,  {P(X > 1 , Y < 1)\over {5 \over 9}}  ,\  \text{and} \ {P(X < 1 , Y < 1)\over {5 \over 9}}$$
$${{1\over 3} \cdot {2\over 3}\over {5 \over 9}} ,  {{2\over 3} \cdot {1\over 3}\over {5 \over 9}}  ,\  \text{and} \ {{1\over 3} \cdot {1\over 3}\over {5 \over 9}}$$
Finally, the expectation of $Y$ is given by integrating and summing the distributions 
$$\int_1^3y{{1\over 3} \cdot {2\over 3}\over {5 \over 9}}dy +  \int_0^1y{{2\over 3} \cdot {1\over 3}\over {5 \over 9}}dy  +  \int_0^1y{{1\over 3} \cdot {1\over 3}\over {5 \over 9}}dy = 1.9$$
However, the correct solution is $1.1$, what is wrong with my approach?

In your first case the conditional expectation is obviously $2$, in the second $\frac12$ and in the third $\frac12$, making the overall conditional expectation $$2\cdot{{1\over 3} \cdot {2\over 3}\over {5 \over 9}} +  \frac12\cdot{{2\over 3} \cdot {1\over 3}\over {5 \over 9}} +  \frac12\cdot{{1\over 3} \cdot {1\over 3}\over {5 \over 9}} = 1.1 $$
In more detail, as BGM says in a comment, your error was in not using the conditional density in your integrals.  In the first case the conditional distribution is uniform over $[1,3]$ so the density is $\frac{1}{3-1}=\frac12$.  In the second and third cases the conditional distribution is uniform over $[0,1]$ so the density is $\frac{1}{1-0}=1$.  So your expression should have been   $$\int_1^3 y\cdot \frac12 \, dy  \cdot {{1\over 3} \cdot {2\over 3}\over {5 \over 9}} +  \int_0^1 y\cdot 1\, dy \cdot {{2\over 3} \cdot {1\over 3}\over {5 \over 9}} +  \int_0^1 y\cdot 1 \, dy  \cdot {{1\over 3} \cdot {1\over 3}\over {5 \over 9}} = 1.1$$ where each integral gives the conditional expectation for that case
