Q:

Product of weak Hausdorff space is weak Hausdorff

I have read on May's Algebraic Topology such that the category of weak Hausdorff space $\mathcal{wTop}$ has same limit as $\mathcal{Top}$, which means

Product of weak Hausdorff space is weak Hausdorff.

recall that:

$X$ is weak Hausdorff means for all $C$ compact Hausdorff and $g: C\to X$, $g(X)$ is closed

this is my idea:
for $f: C\to \prod W_i $ where $C$ is compact and Hausdorff, $W_i$ is weak Hausdorff, 
we proof that every point not in the image has a open neighborhood.
for a point $x$ with coordinates $x_i$.
because WH space is $T_1$ space, points are closed, the subspace $S_i = (x_1,\dots, X_i,x_{i+1},\dots)$ is closed, so $K_i = f^{-1}(S_i)$ is closed, so compact Hausdorff, so we have induced a continuous map from $K_i\to X_i$, so we know the image of this map is closed, so we have open neighborhood of $x$, which is of the form $(X_1, \dots, O_i, X_{i+1},\dots)$. we can proof that (?) we need only finite intersection of this kind of neighborhood to exclude all points in $\text{im } f$, because $C$ is compact. so we are done

Am I correct? I feel this proof is so tedious, any good idea?

EDIT I think now it is correct.
for any two point in $y\in W$, we pick two open neighbourhood, of form $x\in U_x =  (X_1, \dots, O_x, X_{i+1},\dots)$ and $y\in U_y (X_1, \dots, O_y, X_{i+1},\dots)$ where $O_x$ and $O_y$ disjoint. projecting $U_y$s back into $C$, then make it a finite cover. the corresponding finite intersections of $U_x$ will exclude all points

A:

I'm not sure I follow your proof. My approach would be more along the following lines:
Suppose $f: C \rightarrow \prod_{i \in I} X_i$ is continuous, where all $X_i$ are weak Hausdorff, and $C$ is compact Hausdorff. 
Define $C_i = \pi_i[f[C]]$, which are compact and closed subspaces of $X_i$. The maps $f_i:= \pi_i \circ f$ from $C$ onto $C_i$ are closed (as closed subsets of $C$ are themselves compact Hausdorff) and are moreover also perfect (fibres are compact (Hausdorff)). 
The product of the $f_i$ from $C^I$ to $\prod_{i \in I} C_i$ is also perfect (well-known theorem) and so closed in particular. Then compose the diagonal map $e$ (also closed and continuous!) from $C$ into $C^I$ with this product map, and note that this is just $f$ again. I think that should do it.

