Connected Components of the Homotopy Pullback Let $K$, $K'$ and $L$ be simplicial sets and

be a homotopy pullback square. Question: I think that in this situation, the natural map $\pi_0 (L') \rightarrow \pi_0(K') \times_{\pi_0(K)} \pi_0(L)$ is always surjective - is this true?
By the question here, it is certainly now always a bijection, but it seems to me (following my geometric intuition on topological spaces) that this should hold - however I haven't been able to prove it yet. There is probably/ hopefully just an abstract argument that I am not seeing here, however I tried working e.g. with Mayer-Vietoris-sequences and couldn't figure it out, so I hope somone can help me a bit (or give a counterexample...).
You can, if it simplifies things, also assume that our simplicial sets are all Kan complexes, since this case is most interesting to me. Also, it would be interesting to know if we can make similar statements about the higher homotopy groups - but that might be too much to ask for. Anyway, thanks in advance!
 A: We wish to prove the following: given a vertex $x'$ of $K'$ and $y$ of $L$ mapping to the same connected component of $K$, there is at least one vertex $y'$ of $L'$ mapping to the connected component of $x'$ in $K'$ and the connected component of $y$ in $L$.
For simplicity, let me assume that the diagram is a pullback in the ordinary sense, $L \to K$ is a Kan fibration, and $K$ is a Kan complex.
Every homotopy pullback diagram can be replaced with one of this form, so this is no loss of generality.
Let $x$ be the image of $x'$ in $K$.
By hypothesis, the image of $y$ in $K$ is in the same connected component as $x$, so there is an edge of $K$ connecting $x$ to the image of $y$.
Since $L \to K$ is a Kan fibration, we may use the right lifting property with respect to $\Lambda^1_0 \hookrightarrow \Delta^1$ to obtain an edge of $L$ lying over the said edge of $K$.
Thus, we may assume without loss of generality that the image of $y$ in $K$ is precisely $x$.
Since the diagram is a pullback square, this yields a vertex $y'$ of $L'$ mapping to $x'$ in $K'$ and $y$ in $L$.
This proves the claim.
Once you have this basic fact, you get corollaries about homotopy groups almost for free.
The idea is that the homotopy group functor $\pi_n$, considered as a functor from the category of pointed Kan complexes to the category of sets, is the composite of a representable functor (from the simplicially enriched category of pointed Kan complexes to the simplicially enriched category of Kan complexes) and the connected components functor $\pi_0$.
The representable functor preserves homotopy limits, and we already know $\pi_0$ sends homotopy pullback squares to "weak" pullback squares, so the composite $\pi_n$ has the same property.
