I understand that the homotopy category of (pointed) topological spaces and continuous maps is not complete. Nor is it cocomplete. In particular it neither has all pullbacks nor all pushouts.
What are some simple examples of spans $Z\leftarrow X\rightarrow Y$ and cospans $Z\rightarrow X\leftarrow Y$ that cannot be completed to pushout and pullback squares, respectively?
I'll work in the based category, and consider $S^1$ as $\{z\in\mathbb{C}:|z|=1\}$. Consider the maps $$\text{point}\xleftarrow{}S^1\xrightarrow{f}S^1, $$ where $f(z)=z^2$. Suppose that there is a pushout $P$. We would then have a natural isomorphism $[P,X]=\text{Hom}(\mathbb{Z}/2,\pi_1(X))$. On the other hand, the fibration $$ S^1 = B\mathbb{Z} \xrightarrow{f} S^1 \to \mathbb{R}P^\infty = B(\mathbb{Z}/2) \to \mathbb{C}P^\infty = BS^1 \xrightarrow{Bf} \mathbb{C}P^\infty $$ gives an exact sequence $$ [P,S^1] \to [P,\mathbb{R}P^\infty] \to [P,\mathbb{C}P^\infty]. $$ Now $\pi_1(S^1)=\mathbb{Z}$ and $\pi_1(\mathbb{R}P^\infty)=\mathbb{Z}/2$ and $\pi_1(\mathbb{C}P^\infty)=0$, so if $[P,X]=\text{Hom}(\mathbb{Z}/2,\pi_1(X))$ then we have an exact sequence $0\to\mathbb{Z}/2\to 0$, which is impossible.
Here's a class of counterexamples for the pointed homotopy category of connected CW complexes (so even this restriction does not save you). Let $hCW_{\ast}$ denote this category, and let $\pi_{\bullet} : hCW_{\ast} \to \text{Set}_{\bullet}$ denote the functor taking a pointed CW complex to its homotopy groups $\pi_n$. By Whitehead's theorem, $\pi_{\bullet}$ is a conservative functor: it reflects isomorphisms in the sense that if a morphism $f$ has the property that $\pi_{\bullet}(f)$ is an isomorphism, then $f$ must be an isomorphism. Because $\pi_{\bullet}$ consists of a sequence of representable functors, it also preserves any limits that exist in $hCW_{\ast}$. In particular, $hCW_{\ast}$ has, and $\pi_{\bullet}$ preserves, finite products.
On the other hand, $\pi_{\bullet}$ is not a faithful functor: if $\pi_{\bullet}(f) = \pi_{\bullet}(g)$ it does not follow that $f = g$ in the homotopy category (so Whitehead's theorem for maps fails). On the third hand:
Lemma 1: If $C, D$ are categories with equalizers and $F : C \to D$ is a conservative functor which preserves equalizers, then $F$ is faithful.
Proof. Two parallel morphisms $f, g : c \to d$ in $C$ are equal iff their equalizer, as a map to $c$, is an isomorphism. Since $F$ preserves equalizers and is conservative, it also reflects equalizers, so if $F(f)$ and $F(g)$ have an equalizer in $D$, then $f$ and $g$ have an equalizer in $C$. Moreover, if $F(f) = F(g)$, then $F(f)$ and $F(g)$ have equalizer an isomorphism to $F(c)$, and since $F$ is conservative, $f$ and $g$ must also have equalizer an isomorphism to $c$, and so must be equal. $\Box$
Lemma 2: If $C$ has finite products and pullbacks, then it has equalizers. If $F$ preserves finite products and pullbacks, then it preserves equalizers.
Proof. Suppose $f, g : c \to d$ are two parallel morphisms. Then their equalizer is equivalently the pullback of the diagram $c \xrightarrow{(\text{id}_c, f)} c \times d \xleftarrow {(\text{id}_c, g)} c$. In other words, it's given by the intersection of their "graphs" in $c \times d$. $\Box$
Corollary: $hCW_{\ast}$ does not have pullbacks.
This argument is constructive in the sense that any example of a pair of parallel maps $f, g : X \to Y$ in $hCW_{\ast}$ such that $\pi_{\bullet}(f) = \pi_{\bullet}(g)$ but $f \neq g$ produces an example of a pullback that does not exist in $hCW_{\ast}$, namely the pullback of the diagram
$$X \xrightarrow{(\text{id}_X, f)} X \times Y \xleftarrow{(\text{id}_X, g)} X.$$
There are many examples of such maps; for example, we can take $X$ and $Y$ to have homotopy groups in disjoint degrees. The simplest examples have $X = BG, Y = B^2 A$ for some group $G$ and some abelian group $A$; then homotopy classes of maps $X \to Y$ are given by group cohomology classes in $H^2(G, A)$, or equivalently by central extensions of $G$ by $A$. But $\pi_{\bullet}(f) = 0$ for any map $f : X \to Y$.
Here's an example of missing pullback in the homotopy category of spaces, related to Neil's and Qiaochu's answers. I'll work in pointed spaces, though we'll see that the argument also works in unpointed spaces. Let's say we work in the usual homotopy category (i.e., the localization with respect to weak homotopy equivalences).
Assume that $$\require{AMScd} \begin{CD} P @>>> K(\mathbb{Z},2)\\ @VVV @VV {n} V\\ \ast @>>> K(\mathbb{Z},2)\\ \end{CD}$$ is a pullback in the homotopy category. This says that maps $[X,P]$ consist of the subset of $[X,K(\mathbb{Z},2)]$ of maps becoming null-homotopic after composing with $n \colon K(\mathbb{Z},2) \to K(\mathbb{Z},2)$, in other words, the $n$-torsion in $[X,K(\mathbb{Z},2)]$.
Taking $X = S^0$ yields that $P$ is path-connected. Taking spheres $X = S^i$ yields that $P$ is weakly contractible. In particular, $P$ has the singular cohomology of a point. Now let $\alpha \colon K(\mathbb{Z}/n,1) \to K(\mathbb{Z},2)$ be a generator of $H^2( K(\mathbb{Z}/n,1); \mathbb{Z}) = \mathbb{Z}/n$. The equality $n \alpha = 0$ implies that $\alpha$ factors through $P$. This is a contradiction, since $\alpha$ induces a non-zero map on cohomology $H^2(-;\mathbb{Z})$.
Remark 1. The argument also works in the naive homotopy category (i.e., spaces and homotopy classes of maps). The space $P$ might not have the homotopy type of a CW complex and we don't know that singular cohomology $H^2(P;\mathbb{Z})$ is represented by maps $P \to K(\mathbb{Z},2)$; we only used $H^2(P;\mathbb{Z}) = 0$. For the fact that a weak homotopy equivalence induces an isomorphism on singular (co)homology, see Hatcher Proposition 4.21.
Remark 2: Let's adapt the example to unpointed spaces. Recall that for path-connected $Y$, the forgetful map $[X,Y]_* \to [X,Y]$ from pointed homotopy classes of pointed maps to unpointed homotopy classes of unpointed maps is surjective and induces a bijection $[X,Y] = [X,Y]_* / \pi_1(Y)$ to the set of $\pi_1(Y)$-orbits in $[X,Y]_*$; see Hatcher Proposition 4A.2. In the example above, taking $X = \ast$ yields that $P$ is path-connected. Taking $X = S^1$ yields that $\pi_1(P)$ has only one conjugacy class, hence is the trivial group. The rest of the argument goes through.
