$\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}\def\PP{\mathbb{P}}$I noticed these excellent answers were missing an explicit example of a space $X$ and a class $\eta$ in $H_{\ast}(X, \ZZ)$ such that the image of $\eta$ in $H_{\ast}(X \times X)$ (under the diagonal map) is not in the image of $H_{\ast}(X) \otimes H_{\ast}(X) \to H_{\ast}(X \times X)$. So, for the record, this occurs for the fundamental class of $\mathbb{RP}^3$. 

<b>Verification</b> We write points of $\RR \PP^3$ in homogenous coordinates as $(w:x:y:z)$. For $t \in [0,1]$, define 
$$D_t := {\Big \{} ((w_1:x_1:y_1:z_1), (w_2:x_2:y_2:z_2)) \in \RR \PP^3 \times \RR \PP^3 : $$
$$w_2 x_1 = t w_1 x_2,\ x_2 y_1 = t x_1 y_2,\ y_2 z_1 = t y_1 z_2,$$
$$w_2 y_1 = t^2 w_1 y_2,\ x_2 z_1 = t^2 x_1 z_2,\ w_2 z_1 = t^3 z_1 w_2 {\Big \}}$$
and set $D = \bigcup_{t \in [0,1]} D_t$. 

$D_1$ is the diagonal and, for $t \in (0,1]$, we have $D_t \cong \RR \PP^3$, the equations defining $D_t$ simply say that $(w_2:x_2:y_2:z_2) = (t^3 w_1: t^2 x_1: t y_1: z_1)$. The space $D_0$ is a bit more interesting: the equations are $w_2 x_1 = x_2 y_1 = y_2 z_1 = w_2 y_1 = z_2 x_1 = z_2 w_1=0$, and considering cases shows that
$$D_0 = (\RR \PP^3 \times \RR \PP^0) \cup (\RR \PP^2 \times \RR \PP^1) \cup (\RR \PP^1 \times \RR \PP^2) \cup (\RR \PP^0 \times \RR \PP^3).$$
Here $\RR \PP^2 \times \RR \PP^1$, for example, is all points of the form $(\ast: \ast: \ast : 0) \times (0 : 0 : \ast : \ast)$. 

$D$ is an oriented manifold with corners, whose boundary is $D_1 - D_0$. So $D_1$ is homologous to $D_0$. 

Put the standard CW-structure on $\RR \PP^3$, writing $C_i$ for the cell of dimension $i$, and write $C_{ij}$ for $C_i \times C_j$. It is easy to compute in this structure that 
$$H_3 = \ZZ \cdot [C_{30}] \oplus \ZZ \cdot [C_{03}] \oplus (\ZZ/2\ZZ) \cdot ([C_{21}]+[C_{12}]).$$
The image of $\bigoplus_{i+j =3 } H_i(\RR \PP^3) \otimes H_j(\RR \PP^3)$ is the first two summands, and $[D_0] = [C_{30}] + [C_{21}] + [C_{12}] + [C_{03}]$, so $[D_0]$ is not in the image of $\bigoplus_{i+j =3 } H_i \otimes H_j$. (Recall that $H_1(\RR \PP^3) = H_2(\RR \PP^3) = 0$. )

<hr>

I'll point out two things which confused me at first. First of all, why isn't $[D_0]$ the image of $\sum_{i+j=3} [\RR \PP^i] \otimes [\RR \PP^j]$? The answer is that $\RR\PP^2$ is not orientable, so it does not define a class in $H_2(\RR \PP^3)$.

Once we realize that, what does $[D_0]$ mean, since the $\RR \PP^1 \times \RR \PP^2$ and $\RR \PP^2 \times \RR \PP^1$ components are also not orientable? The answer is that the orientation on the interior of $D$ (which is $(0,1) \times \RR \PP^3$, so orientable) gives rise to orientations on the interiors of the components of $D_0$, but these orientations are discontinuous across the codimension $2$ strata where these components meet. Nonetheless, this is enough to give us a map from an oriented closed cell to each component, whose boundaries cancel out, so we can speak of the $H_3$ class of $[D_0]$ even though we can't speak of the classes of its individual components.