Let $f:\mathbb R^n \rightarrow \mathbb R$ be a convex (continuous ) function. Let's assume that some sublevel set $L_a:=\{x\in \mathbb R^n : f(x) \leq a\}$, where $a \in \mathbb R$, contains a halfline $\{p+tu: t\geq 0\}$ with an initial point $p\in L_a$ and a direction $u\in \mathbb R^n$.
How to show that if $b\in \mathbb R$, $b<a$, $ L_b \neq \emptyset$ and $q\in L_b$ then also a halfline $\{q+tu: t\geq 0\}$ (with the same direction $u$ as previous) is contained in $L_b$ ?
I think this is best explained by a picture. The points $P$ and $Q$ are as in your post. The two rays are parallel. We are trying to show that $f(C)\le f(Q)$, given that $f$ is bounded on the ray beginning at $P$.
Using continuity, take a small neighborhood of $Q$ in which $f\le f(Q)+\epsilon$. Draw a line through $C$ which passes through a point in this neighborhood and also meets the ray out of $P$. By convexity, $$ f(C)\le \frac{|AC|}{|AB|}f(B) + \frac{|BC|}{|AB|}f(A) \tag1 $$ Here $f(A)$ is uniformly bounded, and the fraction $\frac{|BC|}{|AB|}$ can be made as small as we wish by moving $A$ out along the ray. Thus, by taking $A$ far away and $B$ close to $Q$, we get $f(C)$ bounded above by a number arbitrarily close to $f(Q)$.