A little abstraction helps here, I think. (Note that my discussion assumes that the feasible set is non-empty. This is true in your example.)
Consider the problem $\max \{d^T x | a_i^T x \leq b_i, \ \ i = 1,...,n \}$.
Suppose you are at some feasible point $x$, and for some direction $h$ the following conditions are satisfied: (1) $a_i^T h \leq 0$ for all $i$, and (2) $d^Th >0$. By considering the point $x(t) = x+th$, you can see that the objective can be increased arbitrarily while keeping the constraints satisfied.
Consequently, to have a bounded maximum, it must be the case (ie, a necessary condition) that if $a_i^T h \leq 0$ for all $i$, then $d^Th \leq0$. This is an important point. The remainder of the answer is showing the connection between this condition and the non-negativity you asked about.
First, a small digression to provide some intuition. Suppose you have two closed linear subspaces $A,B$. Then you have $A \subset B$ iff $B^\bot \subset A^\bot$. In my opinion, this duality (with suitable generalizations) is the essence of understanding and obtaining useful geometric characterizations of many optimality conditions.
Now consider a non-empty closed convex cone $K$. Notations and definitions vary a bit here, but I will define the polar of $K$ to be $K^\bot = \{h | x^T h \leq 0, \ \forall x \in K \}$. The definition matches that of orthogonal complement for subspaces. The essential point here is that if you have two non-empty closed convex cones $K_1, K_2$, then $K_1 \subset K_2$ iff $K_2^\bot \subset K_1^\bot$. You see the formal similarity with subspaces here.
Now return to our problem. Let $A = \mathbb{cone} \{a_i\} = \{ \sum \lambda_i a_i | \lambda_i \geq 0 \}$. $A$ is a non-empty convex cone. A point $x \in A$ iff $x$ can be written as a non-negative combination of the $a_i$. Let $D = \mathbb{cone} \{d \} = \{\lambda d\}_{\lambda \geq 0}$. Then we can rephrase the condition above as if the problem is to have a bounded maximum, we must have $A^\bot \subset D^\bot$. However, this is equivalent to $D \subset A$ by duality, from which we obtain that $d = \sum_i \lambda_i a_i$ for some non-negative constants $\lambda_i$.
In your example, $a_1 = (1,1)^T, a_2 = (-1,0)^T$. Hence to have a bounded maximum, $d = (d_1,d_2)$ must satisfy $d_1\leq d_2$, $d_2 \geq 0$.
Now notice that if we can write $d = \sum_i \lambda_i a_i$ with $\lambda_i \geq 0$, then it follows that the maximum is bounded, because $d^T x = \sum_i \lambda_i a_i^T x \leq \sum_i \lambda_i b_i$.
This shows that in your example, since you have a point $(2,3)^T$ for which all constraints are active, then $a_1^T (2,3)^T = 5$ and $a_2^T (2,3)^T = -2$, then the maximum will be attained at that point and the maximum is given by $5 \lambda_1 - 2 \lambda_2$.
Best Answer
I don't find any mistake in your solution. The feasible region is correctly determined. Of course, the origin is not a feasible solution here.
Sometimes the solution manuals are incorrect. So, don't worry.