the product $(X_1, T_1)\times\ldots\times(X_n,T_n)$ is locally compact, then each $(X_i, T_i)$ is locally compact. Prove that the product $(X_1, T_1)\times\ldots\times(X_n, T_n)$ is locally compact, then each $(X_i, T_i)$ is locally compact.
My proof: Let $p=\langle x_1,\ldots,x_n\rangle$ be in the product.
Since the product is locally compact, there exists a compact neighborhood $K_1\times\ldots\times K_n$ of $p$. Since $K_1\times\ldots\times K_n$ is compact, each factor $K_i$ is compact. 
There is an open set $U_1\times\ldots\times U_n$ in the product such that $p$ is in $U_1\times\ldots\times U_n$ which is the subset of $K_1\times\ldots\times K_n$. So, each $U_i$ is open in $X_i$. For each $x_i$ of a compact neighborhood $K_i$, there exists an open set $U_i$ for $i=1,2,\ldots,n$ such that $x_i$ is in $U_i$ which is the subset of $K_i$. Therefore each $(X_i,T_i)$ is locally compact. Is it ok?
 A: You have most of the pieces, but not quite all of them. Also, you could organize the argument a little better: to show that one of the factor spaces $X_k$ is locally compact, you should start with an arbitrary point of $X_k$ and show that it has a compact nbhd.
Let $X=X_1\times\ldots\times X_n$. Fix $k\in\{1,\ldots,n\}$; you want to show that $X_k$ is locally compact, so let $x_k\in X_k$. For each $i\in\{1,\ldots,n\}\setminus\{k\}$ let $x_i\in X_i$, and let $p=\langle x_1,\ldots,x_n\rangle$. The product space is locally compact, so $p$ has a compact nbhd $K$. Since $K$ is a nbhd of $p$, there are open sets $U_i\in T_i$ for $i=1,\ldots,n$ such that $p\in U_1\times\ldots\times U_n\subseteq K$. For $i=1,\ldots,n$ let $K_i=\operatorname{cl}_{X_i}U_i$; then
$$K_1\times\ldots\times K_n=\operatorname{cl}_X(U_1\times\ldots\times U_n)\subseteq K\;.$$
Thus, $K_1\times\ldots\times K_n$ is a compact nbhd of $p$, and it follows that $K_k$ is compact. But $x_k\in U_k\subseteq K_k$, wher
A: There are competing definitions of local compactness. One of them means that the space has arbitrary small compact neighborhoods. The other one means that the space is Hausdorff and each point has a compact neighborhood. The second one implies the first one.
No matter which one you are using, if the product has the property, then each factor has it. This can be seen by noting that the projection maps are open and continuous. And if $f:X\to Y$ is a continuous open surjection, $y\in Y$ is a point and $x\in X$ such that $f(x)=y$, then any compact neighborhood $K$ of $x$ maps to a neighborhood $f(K)$ of $y$ since $f$ is open and this $f(K)$ is compact since $f$ is continuous. If you want $f(K)$ to be contained in some given neighborhood $V$, just choose $K$ small enough so it is in $f^{-1}(V)$.
Also note that each factor is Hausdorff if the product is. So the statement is true for both definitions.
