Not-too-non-compact metric spaces Consider the following two separable metric spaces: Cantor space $2^\omega$ and Baire space $\omega^\omega$. These spaces give rise to two classes of metric spaces, namely the "big" ones into which Baire space embeds as a closed subset and the "small" ones which are "easily covered" by Cantor space, that is, which are the continuous image of $2^\omega\times\omega$. For example, $\mathbb{R}$ is small in this sense, since $[-n,n]$ is a compact metric space for each $n\in\mathbb{N}$ and every compact metric space is the continuous image of Cantor space. In fact, this last fact shows that "small" is just "$\sigma$-compact."
Each "natural" separable metric space that I can think of is either big or small in this sense, and this raises the following question:

Suppose $M$ is a separable metric space which doesn't contain a closed set homeomorphic to Baire space and is homeomorphic to a Borel subset of Baire space (I'll also accept more complicated examples as long as they are "reasonably natural," e.g. exist in $L(\mathbb{R})$ assuming large cardinals). Must $M$ be $\sigma$-compact? 

I suspect the answer is "no," but I haven't been able to whip up a counterexample.
 A: I think the answer is yes for Borel subsets of Polish spaces. I'm not confident enough to say anything about $L(\mathbb R).$

Lemma 1. Any second-countable space $M$ can be partitioned into an open $\sigma$-compact space $A$ and a closed subspace $B$ such that every non-empty relatively open subset of $B$ is non-$\sigma$-compact.

Proof: Define $A$ to be the union of all $\sigma$-compact open sets in $M.$ Since $M$ is strongly Lindelöf, $A$ is $\sigma$-compact. Set $B=M\setminus A.$
Consider a relatively open $\sigma$-compact subset $U\subseteq B.$ This is the restriction of some open $U'\subseteq M.$ Since $A\cup U'=A\cup U$ is open and $\sigma$-compact, $U'\subseteq A.$ So $U=\emptyset.$ $\Box$
In fact I'm only going to use a weaker property of $B$ as a metric space: closed balls are non-compact. (If a closed ball $B'(x,r)$ is compact, then $B(x,r)$ is $\sigma$-compact.)

Lemma 2. If $M$ is a non-empty subspace of a complete metric space $P,$ and $M$ has the property of Baire, and all closed balls of $M$ are non-compact, then there is a closed embedding of $\omega^\omega$ in $M.$

Proof. Write $M=U\Delta (\bigcup C_i)$ where $U$ is open and each $C_i$ is nowhere dense.
Since $M$ is not sequentially compact, it is either not totally bounded or it has a Cauchy sequence that converges to a point in $P\setminus M.$ In either case we can find 
a sequence of disjoint "non-converging" closed balls $B_n\subseteq U$ - by this I mean that no sequence with $x_n\in B_n$ has limit points in $M.$ We can pick the balls $B_n$ so that they do not intersect $C_1$ and such that they have radius less than $1$.
Then for each $n_1$ we can pick a sequence of disjoint non-converging closed balls $B_{n_1,n_2}\subseteq B_{n_1}$ which do not intersect $C_2$ and with radius less than $1/2.$
Continue in this manner, defining closed balls $B_{n_1,\dots,n_{k+1}}\subseteq B_{n_1,\dots,n_k}$ avoiding $C_1\cup\cdots\cup C_{k+1}$ and with radius less than $1/k.$ 
Define an embedding $f:\omega^\omega\to M$ by taking $(n_1,n_2,\cdots)$ to the unique point in $\bigcap_{k}B_{n_1,\dots,n_k}.$
This map is injective because for each $k,$ the balls $B_{n_1,\dots,n_k}$ are disjoint.
For each sequence $x_1,x_2,\dots\in\omega^\omega,$ the sequence $f(x_n)$ converges if and only if $x_n$ converges in $\omega^\omega.$ So $f$ is a closed embedding. $\Box$
Borel subsets $M$ of a Polish space certainly have the Baire property. So these two lemmas imply that $M$ must be either $\sigma$-compact or contain a homeomorphic copy of $\omega^\omega.$
