Question about almost metrizable space and $k$-spaces. 
A space $X$ is said almost metrizable if for every $x\in X$ there exists 
  a compact set $F \subset X$ with a countable neighbourhood base in $X$ such that $x\in F$.

Is every almost metrizable space $q$-space? How about $k$-spaces?

A space $X$ is $q$-space if for every $x\in X$ there exists a sequence $\{U_n : n\in\omega\}$ of open neighbourhoods of $x$ in $X$ such that For every sequence $\xi=\{ x_n : n\in \omega\}$ of points in $X$ such that $x_n\in U_n$ for each 
$n$, there exists a limit point in $X$. 
A Hausdorff space $X$ is a $k$-space if and only if for each $A\subset X$, the set
$A$ is open in $X$ provided that the intersection of $A$ with any compact subspace $Z$ of the space
$X$ is open in $Z$.
 A: Your almost metrizable spaces are called spaces of point-countable type (that's how I knew them); they were introduced by Arhangel'skij, see the first two references in this paper, where it is mentioned that they are $k$-spaces, as proved in the second reference. 
I also think they are $q$-spaces: just use the local base $U_n$ for the compact set $K$ that contains $x$; make it decreasing WLOG. If we pick $x_n \in U_n$, I think we must get a cluster point on $K$ by compactness. BTW this paper in corollary 0.2 claims that this is proved in a Quintuple Quotient Quest by E. Michael (to which I have no access). 
Added from comments To make the answer more self-contained, I'll add the proof as filled in by Alex Ravsky in the comments. My idea from above is indeed correct: suppose no point of $K$ is a cluster point of $\{x_n: n \in \omega\}$, then every point $x \in K$ has an open neighbourhood $O_x$ that only contains at most finitely many $x_n$. The open cover $\{O_x: x \in K\}$ has a finite subcover with union $O$. Then $O$ is a neighbourhood of $K$ that only contains finitely many $x_n$ as well, so from some index $N$ onwards, no $x_n$ with $n \ge N$ will be in $O$. But now we have a contradiction, as for some $k > N$ we will have that $U_k \subset O$ (as we have a decreasing neighbourhood base for $K$ in the $U_n$), and then $x_k \in U_k \subset O$, which cannot be.
The Quintuple Quotient Quest also indirectly shows that a space of pointwise countable type (as an open image of a paracompact $M$-space) is a $k$-space (as a quotient image of a paracompact $M$-space). But as said, this was already known to Arhangel'skij.
