Compact group actions and automatic properness

I am currently re-reading a course on basic algebraic topology, and I am focussing
on the parts that I feel I had very little understanding of. There is one exercise
in the chapter devoted to groups acting on topological spaces (preceding the chapter
on covering spaces) that I have spent several frustrating hours on, but I just can’t
crack.

I will state the question below, but PLEASE, DO NOT POST A SOLUTION OR A HINT. I am
only interested in knowing wether the statement in question is true. I have looked
know a book that gives proves it, or a document online where this is discussed, I
would like to get the reference, in case I continue to fail at giving a proof of
this the next week.

I recall some terminology first: let $X,Y$ be two topological spaces, and
$f:X\rightarrow Y$ a map that isn’t supposed to be continuous. The author defines
such a map to be $\mathrm{PROPER}$ whenever the following two properties are
satisfied : $f$ is a closed map, and $\forall y\in Y, ~f^{-1}(\lbrace y\rbrace)$ is
a compact subspace of $X$. It then follows that for all compact subsets $K$ of
$Y,~f^{-1}(K)$ is a compact subset of $X$. Also, if $X$ is Hausdorff, and $Y$ is a
locally compact Hausdorff, then properness is equivalent to this property.

Let $X$ be a topological space, and $G$ a topological group. Suppose there is a
continuous left group action $\rho: G\times X\rightarrow X,~(g,x)\mapsto g\cdot x$.
Let $\theta:G\times X\rightarrow X\times X, ~(g,x)\mapsto (x,g\cdot x)$. The author
defines the group action to be $\mathrm{proper}$ if $\theta$ is a proper map.

Here is the question: “Show the action of a compact Hausdorff group $G$ on a
Hausdorff space $X$ is always proper”.

As I said, I have struggled with this for days (since friday). IS THAT STATEMENT
TRUE? There are no further hypothesis, $X$ is not supposed to be locally compact,
and the action is completely arbitrary (continuous of course, but not supposed free,
or other things).