This is only a partial answer; but it's too long for a comment, and I believe the following facts, which come close to the question being asked and might provide a clue to answering it, are worth pointing out.
Let me show that there exists a bijection $\mathbb{R} \to \mathbb{R}^2$ such that the forward map is connected but the inverse is not.
Let me call a subset $E$ of the plane "hyperdense" when it meets every perfect set of the plane (perfect = nonempty, closed, with no isolated point). In particular, it meets every nonempty open set and every uncountable closed set (by Cantor-Bendixson). Let me reproduce the argument given here to show that every hyperdense set of the plane is connected: if not, there would be $U,V\subseteq\mathbb{R}^2$ open such that $E \subseteq U\cup V$, $E$ meets both $U$ and $V$ and does not meet $U\cap V$; but since $E$ is hyperdense, not meeting the open set $U\cap V$ is possible only if $U\cap V = \varnothing$, so $U\cup V$ is disconnected, so the closed set $\mathbb{R}^2 \setminus (U\cup V)$ must be uncountable (because the complement of a countable set in the plane is connected), so $E$ must meet it, a contradiction.
Now I construct a bijection $\mathbb{R} \to \mathbb{R}^2$ such that $f(I)$ is hyperdense for every nonempty open interval $I$. Let $\mathfrak{c} = 2^{\aleph_0}$ (seen as the smallest ordinal of that cardinality); let $(I_\beta,P_\beta)_{\beta<\mathfrak{c}}$ be an enumeration of pairs consisting of a nonempty open interval $I \subseteq \mathbb{R}$ and a perfect set $P \subseteq \mathbb{R}^2$ (recall that there are precisely continuum-many perfect sets), and let $(z_\beta)_{\beta<\mathfrak{c}}$ and $(t_\beta)_{\beta<\mathfrak{c}}$ be an enumeration of $\mathbb{R}$ and $\mathbb{R}^2$ respectively. Construct $(x_\alpha,y_\alpha) \in \mathbb{R} \times \mathbb{R}^2$ by induction on $\alpha<\mathfrak{c}$ such that $x_\alpha$ is different from all the (previously constructed) $x_{\alpha'}$ for $\alpha'<\alpha$ and $y_\alpha$ is different from all $y_{\alpha'}$ for $\alpha'<\alpha$, and additionally: (A) if $\alpha=2\beta$ is even, then choose $x_\alpha \in I_\beta$ and $y_\alpha \in P_\beta$ (these conditions can be met because $I_\beta$ and $P_\beta$ have cardinality $2^{\aleph_0}$ and there are $<2^{\aleph_0}$ previously constructed $x_{\alpha'}$ and $y_{\alpha'}$ to be avoided); (B₁) if $\alpha=4\beta+1$ then take $x_\alpha=z_\beta$ unless $z_\beta$ is already among the previous $x_{\alpha'}$ (otherwise, the choice is unconstrained), and (B₂) if $\alpha=4\beta+3$ then take $y_\alpha=t_\beta$ unless $t_\beta$ is already among the previous $y_{\alpha'}$ (otherwise, the choice is unconstrained; conditions (B₁) and (B₂) are only there to guarantee that $f$ will be defined everywhere and surjective). Clearly the choice can be made at every stage, so the sequence $(x_\alpha,y_\alpha)_{\alpha<\mathfrak{c}}$ exists. By construction, the $x_\alpha$ are distinct and by (B₁) every real $z_\beta$ is equal to some $x_\alpha$, so we can define $f(x_\alpha) = y_\alpha$, giving a function $\mathbb{R} \to \mathbb{R}^2$, which is injective because the $y_\alpha$ are distinct and surjective by (B₂). Now if $I$ is any nonempty open interval and $P$ is some perfect set, we can write $(I,P) = (I_\beta,P_\beta)$ and if $\alpha = 2\beta$ we have $f(x_\alpha) = y_\alpha \in P$ for $x_\alpha \in I$, which shows $f(I) \cap P \neq \varnothing$. So $f(I)$ is always hyperdense.
Now if $I \subseteq \mathbb{R}$ is connected, either it is empty or a singleton, in which case $f(I)$ is trivially connected, or it is an interval with nonzero length, but then it contains a nonempty open interval, so $f(I)$ is hyperdense, so it is connected as explained above. This shows that the forward map of $f$ is connected.
On the other hand, the inverse map of $f$ cannot be connected, because if it were, the inverse image of an open disk in $\mathbb{R}^2$ would be connected, hence an interval, which cannot be trivial since $f$ is bijective, so it must contain an open interval, so $f$ is continuous; but $f$ is certainly not continuous since it is bounded on no interval.
In a related vein, let me show that there exists a bijection $f\colon \mathbb{R}^2 \to \mathbb{R}^2$ such that the forward image of every PATH-connected set is connected and such that the inverse map is not connected.
The construction above easily gives a bijection $f\colon \mathbb{R}^2 \setminus\Delta \to \mathbb{R}^2 \setminus Z$, where $\Delta$ is the diagonal and $Z = (0,1)\times\{0\}$, such that $f(Q) \cap P \neq\varnothing$ for every perfect set $Q$ in $\mathbb{R}^2 \setminus\Delta$ and every perfect set $P$ in $\mathbb{R}^2 \setminus Z$, i.e., $f(Q)$ is hyperdense in $\mathbb{R}^2 \setminus Z$ for every perfect $Q$ in $\mathbb{R}^2 \setminus\Delta$.
Extend $f$ to a bijection $\mathbb{R}^2 \to \mathbb{R}^2$ by using a homeomorphism between $\Delta$ and $Z$.
Clearly, the inverse map of $f$ is not connected, since $f^{-1}(\mathbb{R}^2 \setminus Z) = \mathbb{R}^2 \setminus\Delta$ and $\mathbb{R}^2 \setminus\Delta$ is not connected whereas $\mathbb{R}^2 \setminus Z$ is.
Now consider the forward image of a path-connected set $A \subseteq \mathbb{R}^2$. If $A \subseteq \Delta$ then $f(A)$ is connected because $f$ restricts to a homeomorphism between $\Delta$ and $Z$. On the other hand, if $A \not\subseteq \Delta$, clearly $A\setminus\Delta$ is not a single point, so $A$ must contain a path between two points not in $\Delta$, and at least part of this path is not in $\Delta$, so it contains a perfect set $Q \subseteq \mathbb{R}^2 \setminus\Delta$. So $f(Q)$ is hyperdense in $\mathbb{R}^2 \setminus Z$; in particular, it is dense in $\mathbb{R}^2$, and connected (the same argument showing that hyperdense sets in $\mathbb{R}^2$ are connected still works for $\mathbb{R}^2\setminus Z$ since the complement of a countable set in the latter is connected). Now $f(A)$ contains $f(Q)$ which is dense in $\mathbb{R}^2$ connected, and anything containing a dense connected set is still connected (because if $D$ is dense connected and $D \subseteq B$, then any continuous $B\to\{0,1\}$ has a constant restriction to the dense subset $D$ so it is constant). So $f(A)$ is connected.
It is conceivable that the same kind of arguments give a positive answer to the original question, provided we can find something to play the role of the intervals $I$ in the first part and the perfect sets $Q$ in the second. But the naïve approach fails: indeed, one might ask the following question
Question: Does there exists $2^{\aleph_0}$ subsets $(H_\gamma)$ of the plane $\mathbb{R}^2$, each of cardinality $2^{\aleph_0}$, such that every connected $A \subseteq \mathbb{R}^2$ that is not a singleton contains one of the $H_\gamma$?
(Then we could define a bijection $f\colon\mathbb{R}^2\to\mathbb{R}^2$ by transfinite induction as above so that $f(H_\gamma)$ is hyperdense for each $H_\gamma$. Its forward map would be connected, but it wouldn't be continuous so this would answer the "equivalent question" cited in the original post.)
However, the answer to this question must be "no", because if such $(H_\gamma)$ existed, we could also use them to construct a bijection $f\colon\mathbb{R}^2\to\mathbb{R}^2$ such that $f(H_\gamma)$ and $f^{-1}(H_\gamma)$ are both hyperdense for each $H_\gamma$, which would make both the forward and inverse maps of $f$ connected, but $f$ would not be continuous (say we include all perfect sets among the $H_\gamma$ to be sure), contradicting the result of Tanaka cited by Willie Wong's question. (Probably there exists a more direct proof that the question has a negative answer. [EDIT: Will Brian gives such a proof in the comments]) So either this approach is doomed or we must be smarter in how to use it.
Incidentally, does ZFC prove that connected subsets of the plane satisfy the continuum hypothesis? [EDIT: again, Will Brian gave a (positive) answer in the comments]
A counterexample with finite topological spaces:
recall that on a set $X$, an "Alexandroff discrete" topology ($\mathrm T_0$ where all intersections of open sets are open)
is the same thing as a partial order; the minimum open set containing $x$ is the principal order filter $Fx$ in the order.
The dual order gives the topology where open and closed sets are interchanged.
Take a finite $X$, and as $Y$ its dual as above; take as $C$ the principal diagonal of points $(x,x)$.
The minimum open set which contains $(x,x)$ is the direct product of the principal order filter $Fx$ and the principal ideal $Ix$.
The only $y$ with both inequalities $x\leq y$, $x\geq y$ is $y=x$, so the diagonal $C$ is discrete.
Essentially, one has the identity map on a finite set $X$, with the discrete topology in the domain and an arbitrary $\mathrm T_0$ topology
in the codomain. No identifications but no homeomorphism either (except for the antichain poset order on $X$). No quotient.
Edit. Take any two distinct quasi compact topologies on a set $X$,
and the projections from the diagonal as above. At least one
of the two projections is not a quotient.
Best Answer
A counterexample to this question can be constructed as follows.
Let $X=[0,1]$ be the closed interval with the standard Euclidean topology.
Let $Y=\omega$ and $\mathcal T_Y$ be the topology on $Y$ consisting of the sets $W\subseteq Y$ satisfying two conditions:
$\bullet$ if $0\in W$, then $W=\omega$;
$\bullet$ if $1\in W$, then $\omega\setminus W$ is finite.
The definition of the topology $\mathcal T_Y$ implies that $\omega\setminus\{0,1\}$ is an open discrete subspace of $(Y,\mathcal T_Y)$.
It is easy to see that the space $(Y,\mathcal T_Y)$ is not locally connected at $1$.
Choose any sequence $(U_n)_{n=1}^\infty$ of pairwise disjoint nonempty open sets in $X=[0,1]$ and let $g:X\to Y$ be the function defined by $$g(x)=\begin{cases}n&\mbox{if $x\in U_n$ for some $n\ge 1$};\\ 0&\mbox{otherwise}. \end{cases} $$ The definition of the topology $\mathcal T_Y$ ensures that the function $g:X\to Y$ is continuous.
Let $C=\{(x,g(x)):x\in X\}\subseteq X\times Y$ be the graph of the function $g$. It is clear that $C$ is homeomorphic to $[0,1]$ and hence is compact, connected and locally connected. But the projection of $C$ onto $Y$ is not locally connected.