ThisLet us recall that a family $\mathcal N$ of subsets of a topological space $X$ is onlycalled a partial answer$cs$-network if for any sequence $\{x_n\}_{n\in\omega}\subset X$ that converges to a point $x\in X$ and any neighborhood $U\subset X$ of $x$ there exists a set $N\in\mathcal N$ such that $x\in N\subset U$ and $x_n\in N$ for all but finitely many numbers $n\in\omega$.
Given a classThe following theorem answers the problem posed by Sam Nolen. I thank Martin Sleziak for his comment containing the reference to $\mathcal C$the paper of Franklin and Rajagopalan.
Theorem. The smallest class of topological spaces that contains $\mathbb R$ and is closed under taking subspaces and quotient spaces coincides with the class of subspaces of sequential spaces that have cardinality $\le \mathfrak c$ and possess a countable $cs$-network.
Proof. We shall prove that $\mathcal R=S\mathcal K=SQSQS\mathbb R$, where
$\bullet$ $\mathcal R$ is the smallest class of topological spaces, let $S\mathcal C$ and $Q\mathcal C$ be respectivelywhich contains the classes ofreal line and is closed under taking subspaces and quotient spacesspaces;
$\bullet$ $\mathcal K$ is the class of sequential spaces of cardinality $\le\mathfrak c$ that belong topossess a countable $cs$-network;
$\bullet$ $S\mathcal K$ is the class of subspaces of spaces in the class $\mathcal C$$\mathcal K$.
Let$\bullet$ $S\mathbb R$ is the family of subspaces of the real line;
$\bullet$ $QS\mathbb R$ beis the class of quotient spaces of spaces in the class $S\mathbb R$;
$\bullet$ $SQS\mathbb R$ is the class of subspaces of spaces in the class $\mathbb R$ and$QS\mathbb R$;
$\bullet$ $(QS)^{n+1}\mathbb R:=QS(QS^n\mathbb R)$ for$QSQS\mathbb R$ is the class of quotient spaces of spaces in the class $n\in\mathbb N$$SQS\mathbb R$;
$\bullet$ $SQSQS\mathbb R$ is the class of subspaces of spaces in the class $QSQS\mathbb R$;
The proof of the equality $\mathcal R=S\mathcal K=SQSQS\mathbb R$ is divided into six lemmas.
The problem posedfirst lemma was proved in the OP isanswer of Will Brian.
Lemma 1. The class $QS\mathbb R\subset\mathcal R$ contains all separable metrizable spaces.
Lemma 2. For any zero-dimensional Polish space $Z$ and any anti-discrete space $A$ of cardinality continuum the product $Z\times A$ belongs to describe the class $QS^\omega\mathbb R:=\bigcup_{n\in\mathbb N}QS^n\mathbb R$$QS\mathbb R\subset\mathcal R$.
I was ableProof. The anti-discrete space $A$ can be identified with the topological group $\mathbb R/\mathbb Q$ endowed with the quotient topology (which is anti-discrete). Then $Z\times A$ belongs to trace$QS\mathbb R$, being a quotient space of the separable metrizable space $Z\times \mathbb R$, which belongs to $QS\mathbb R$ by Lemma 1.
Lemma 3. $\mathcal K\subset QSQS\mathbb R\subset \mathcal R$.
Proof. Let $X$ be a sequential space of cardinality $\le\mathfrak c$ that has a countable $cs$-network $\mathcal N$.
Without loss of generality, we can assume that $\mathcal N$ is closed under finite unions and intersections. Endow the classesset $QS^n\mathbb R$ only$\mathcal N$ with the discrete topology and consider the zero-dimensional Polish space $\mathcal N^\omega$.
Let $X_a$ be the set $X$ endowed with the anti-discrete topology $\{\emptyset, X\}$.
In the space $\mathcal N^\omega\times\ X_a$ consider the subspace
$Z$ consisting of pairs $((N_k)_{k\in\omega},x)\in\mathcal N^\omega\times X_a$ such that for smallany open neighborhood $n$$U\subset X$ of $x$ there exists $n\in\omega$ such that $x\in N_{k+1}\subset N_k\subset U$ for all $k\ge n$.
By Lemma 2, the space $Z$ belongs to the class $SQS\mathbb R$.
NamelyLet $q:Z\to X$, $QS\mathbb R$$q:((N_k)_{k\in\omega},x)\mapsto x$, be the projection on the second factor. It is easy to see that the class ofmap $q$ is surjective and continuous. To see that $q$ is quotient, we need to check that a set $F\subset X$ is closed if its preimage $q^{-1}(F)$ is closed in $Z$. Assuming that $F$ is not closed in the sequential spaces withspace $X$, we can find a sequence $\{x_n\}_{n\in\omega}\subset F$ that converges to some point $x\in X\setminus F$. Let $\mathcal N'$ be the family of all sets $N\in\mathcal N$ such that $x\in N$ and $x_n\in X$ for all but finitely many numbers $n$. Let $(N_i')_{i\in\omega}$ be an enumeration of the countable set $\mathcal N'$. For every $k\in\omega$ let $N_k=\bigcap_{i\le k}N_i'$.
Taking into account that the $k$-network (this$\mathcal N$ is closed under finite unions, we can be derived from Michael's Theorem 11show that $z:=((N_k)_{k\in\omega},x)\in Z$.3 in It is clear that $z\notin q^{-1}(F)$. It can be shown that the Gruenhage's survey article inpair $z$ belongs to the closure of Handbook of Set-Theoretic Topology)$q^{-1}(F)$, which is not possible as $q^{-1}(F)$ is closed in $Z$.
This contradiction shows that $X$ is a quotient space of $Z$ and hence $X\in QSQS\mathbb R$.
Lemma 3 implies
Lemma 4. $S\mathcal K\subset SQSQS\mathbb R\subset \mathcal R$.
Lemma 5. The class $S\mathcal K$ is closed under quotient spaces.
Proof. Let $S$ be a subspace of a space $K\in\mathcal K$ and $q:S\to X$ be a quotient map of $S$ onto some topological space $X$. The map $q$ determines the equivalence relation $$E=\{(x,y)\in K\times K:x=y\}\cup\{(x,y)\in X\times X:q(x)=q(y)\}.$$ In Proposition 3.2 of this paper, Franklin and Rajagopalan prove that $X$ can be identified with a subspace of the quotient space $K/E$.
The following lemma ensures that the quotient space $K/E$ belongs to the class $SQS\mathbb R$ need not$\mathcal K$.
Lemma 6. Let $f:X\to Y$ be sequentiala quotient map. If (but still have$X$ is a sequential space with countable $k$$cs$-network), then so is the space $Y$.
How to describeProof. The sequentiality of the classquotient space $QSQS\mathbb R$$Y$ is a misterywell-known fact (for me at the momentthat can be found in Engelking, I hope). So, it remains to prove that the space $Y$ has a countable $cs$-network.
OnLet $\mathcal N$ be a countable $cs$-network for the other handspace $X$. Without loss of generality, we can assume that the classfamily $QS^\omega\mathbb R$$\mathcal N$ is contained inclosed under finite unions.
Consider the classfamily $\mathcal M:=\{q(N):N\in\mathcal N\}$ and for every $M\in\mathcal M$ let $\ddot M$ be the intersection of spaces withall open sets in $Y$ than contain $M$. We claim that the countable networkfamily (which$\ddot{\mathcal M}=\{\ddot M:M\in\mathcal M\}$ is preserved by taking subapcesa $cs$-network for the space $Y$. Since the family $\mathcal N$ is closed under finite unions, so are the families $\mathcal M$ and continuous images)$\ddot{\mathcal M}$.
It isFix a good questionsequence $\{y_n\}_{n\in\omega}\subset Y$, convergent to find a spacepoint $y\in Y$ and let $U\subset Y$ be a neighborhood of $y$ in $Y$. Let $\mathcal M'=\{M\in\ddot{\mathcal M}: M\subset U\}$. We claim that the family $\mathcal M'$ contains a set $M$ with $y\in M$. To find such set $M$, take any point $x\in q^{-1}(y)$ and find a set $N\in\mathcal N$ such that $x\in N\subset q^{-1}(U)$. Then $M=q(N)$ contains $y$ and the set $\ddot M$ contains $y$ and belongs to the family $\mathcal M'$.
So, we can choose an enumeration $\{\ddot M_k\}_{k\in\omega}$ of the countable networkfamily $\mathcal M'$ such that $y\in \ddot M_0$. We claim that for some $k\in\omega$ the set $\bigcup_{i\le k}\ddot M_i$ contains all but finitely many points $y_n$. Assuming that this is not true, we can construct an increasing number sequence $(n_k)_{k\in\omega}$ such that $y_{n_k}\notin \bigcup_{i\le k}\ddot M_i$. Taking into account that the set $\ddot M_0$ contains $y$, but $\ddot M_0$ does not contain the points $y_{n_k}$, we conclude that $y$ does not belong to the classclosure $QS^\omega\mathbb R$$\overline{\{y_{n_k}\}}$ of the singleton $\{y_{n_k}\}$ for all $k\in\omega$. Consequently, the set $B=(X\setminus U)\cup\bigcup_{k\in\omega}\overline{\{y_{n_k}\}}$ does not contain its accumulation point $y$ and hence is not closed in $Y$.
MaybeSince the map $C_p[0,1]$?$q$ is quotient, the preimage $q^{-1}(B)$ is not closed in $X$. By the sequentiality of $X$, there exists a sequence $(x_m)_{m\in\omega}\in q_E^{-1}(B)$ that converges to some point $x\notin q^{-1}(B)$. It follows from $X\setminus q^{-1}(U)=q^{-1}(Y\setminus U)\subset q^{-1}(B)$ that $x\in q^{-1}(U)$.
By the definition of a $cs$-network, there exists a set $N\in\mathcal N$ such that $x\in N\subset q^{-1}(U)$ and $N$ contains all but finitely many points $x_m$. Find $k\in\omega$ such that $q(N)=M_k$. Since the closed set $F_k:=\bigcup_{i\le k}q^{-1}(\overline{\{y_{n_i}\}})$ does not contain $x$, it does not contain the points $x_m$ for all sufficiently large numbers $m$. So, we can find $m\in\omega$ so large that $x_m\notin F_k$. Then $x_m\in q^{-1}(\overline{\{y_{n_i}\}})$ for some $i>k$ and hence $q(x_m)\in \overline{\{y_{n_i}\}}$. Now observe that $q(x_m)\in q(N)=M_k$ and hence $y_{n_i}\in\ddot M_k\subset \bigcup_{j\le i}\ddot M_j$, which contradicts the choice of $y_{n_i}$.
