Connected preimage of quotient map I'm reading a paper on the fundamental groups of quotient spaces, and thought of the following question: 
Let $f: X \to Y$ be a quotient map with $X$ locally path connected and path connected, and further assume that each fiber $f^{-1}(y)$ is connected. Is $f^{-1}(U)$ connected for each connected subset $U\subset Y$?
Any ideas would be appreciated!
 A: Here's a counterexample.
Let $X=\{a,b,c,d\}$ with the topology having the base $\{a,b\},\{b\},\{b,c,d\},\{d\}$, and let $Y=\{A,C,D\}$ with the topology generated by $\{C,D\},\{D\}$. Then 
$$ q:X\to Y, \quad a\mapsto A,\; b,c\mapsto C,\; d\mapsto D $$
is a quotient map with connected fibers. However, $U=\{A,D\}$ is a connected set in $Y$ with disconnected preimage $\{a,d\}$
A: Just a remark: the assertion is true if $U\subset Y$ is assumed open or closed.
No connectedness assumptions are needed on $X$.
Added. Pedro's answer with additional details made me realize my answer might be fairly useless, so I thought to add my viewpoint as a series of exercises.
Proposition 1. Let $X_0\subset X$ be clopen and $C\subset X$ connected. Then $C\cap X_0\neq\emptyset\implies C\subset X_0$.
Definition 1. A continuous map is monotone if its non-empty fibers are connected.
Proposition 2. Suppose $f:X\to Y$ is monotone. If $X_0\subset X$ is clopen then $X_0=f^{-1}fX_0$.
Definition 2. A continuous map $f:X\to Y$ is a quotient map if $V\subset Y$ is open iff $f^{-1}V\subset X$ is open. (We won't need surjectivity.)
Proposition 3. Suppose $f:X\to Y$ is a quotient map. If $f$ is monotone and $Y$ is connected then $f$ is surjective and $X$ is connected.
Proposition 4. Suppose $f:X\to Y$ is a quotient map. If $B\subset Y$ is open or closed then $f:f^{-1}B\to B$ is a quotient map.
Combining everything gives the assertion in the first sentence of the answer. Here's a neat corollary of the third proposition.
Corollary 1. A finite product of connected spaces is connected.
Proof. Let $X,Y$ be connected and consider the projection map $X\times Y\to Y$. It's a quotient map and the fibers are connected because they're homeomorphic to $X$. Since $Y$ is connected we may apply the proposition to deduce $X\times Y$ is connected.
A: I had to think about this just now, so let me elaborate a bit on Arrow's answer for the sake of completeness.

Let $f\colon X\to Y$ be a continuous, closed and surjective function with non-empty connected fibers.
  Let $Z\subseteq Y$ be a non-empty closed and connected subset.
  Then $f^{-1}(Z)$ is connected.

We can check this with the following characterization:

A topological space is connected if and only if every non-empty closed and open subset is the whole space.

So let $\varnothing \neq A\subseteq f^{-1}(Z)$ be a closed and open subset.
The goal is to show that $A=f^{-1}(Z)$.


*

*$f(A)$ is a non-empty subset of $Z$.

*$f(A)$ is a closed subset of $Z$. Indeed, since $Z$ is closed and $f$ is continuous, $f^{-1}(Z)$ is also closed. Hence $A$ is closed in $X$.
Since $f$ is closed, $f(A)$ is closed in $Y$, thus also in $Z$.

*$f(f^{-1}(Z)\setminus A)=Z\setminus f(A)$. Since $f$ is surjective, we have $Z\setminus f(A)\subseteq f(f^{-1}(Z)\setminus A)$. Let now $z=f(x)$ for $x\in f^{-1}(Z)\setminus A$. Then $x\in f^{-1}(z)\setminus A\cap f^{-1}(z)$, so $A\cap f^{-1}(z)\subsetneq f^{-1}(z)$ is a proper closed and open subspace of the fiber. Since the fiber is connected, it follows that $A\cap f^{-1}(z)=\varnothing$, which means that $z\in Z\setminus f(A)$.

*$f(A)$ is an open subset of $Z$. Since $f^{-1}(Z)$ is closed in $X$ and $A$ is open in $X$, $f^{-1}(Z)\setminus A$ is closed in $X$. Since $f$ is closed, $Z\setminus f(A)$ is closed in $Y$ by the previous point, thus also in $Z$.

*$f(A)=Z$. This follows from $Z$ being connected and points 1, 2 and 4 above.

*$f^{-1}(f(A))=A$. We always have $A\subseteq f^{-1}(f(A))$, so let us check the other inclusion. Let $x\in f^{-1}(f(A))$, so that there is an $a\in A$ with $f(x)=f(a)$. This implies that $\varnothing \neq A\cap f^{-1}(f(x))\subseteq f^{-1}(f(x))$ is a non-empty closed and open subset. Since the fiber is connected, this implies that $A\cap f^{-1}(f(x))=f^{-1}(f(x))$, hence $f^{-1}(f(x))\subseteq A$ and $x\in A$ as well.

*$A=f^{-1}(Z)$, which is what we wanted to show. This follows from points 5 and 6 above.
Side note:
In case anyone reading this is interested in algebraic geometry, we can combine this with Hartshorne's version of Zariski's Main Theorem [Corollary III.11.4 in his Algebraic Geometry book] to obtain the following statement:

Let $f\colon X\to Y$ be a proper birational morphism of (irreducible) algebraic varieties. Assume that $Y$ is normal. Then the preimage of every connected closed subset of $Y$ is again connected.

