**Edit 1.a:**This construction tried to provide a counterexample, but is wrong. I give more details below.

Glue two solenoids as follows.

Let $K$ be a Cantor set (e.g. iterate $[0,1]\to [0,\frac 1 3] \cup [\frac 2 3,1]$). Let $D\subset K$ be a closed nowhere dense Cantor subset (e.g. iterate $[0,1]\to [0,\frac 1 9] \cup [\frac 8 9, 1]$).

Let $S_1,S_2$ be two solenoids over circles $C_1,C_2$ respectively. Essentially, a solenoid is a space $S$ that fibres over a circle $S\to C$, with all the fibres homeomorphic to Cantor sets, and with a certain nontrivial twist. A solenoid is an indecomposable continuum, is homogeneous, and the only proper subcontinua are arcs.

On each circle $C_i$ consider two points $P_i^0, P_i^\pi$, and let $F_i^\alpha$ be the corresponding fibers on $S_i$, for $i=1,2$ and $\alpha=0,\pi$. Let homeomorphisms $a: F_1^0\to D$, $b:K\to F_2^0$, $c: F_2^\pi\to D$, $d:K\to F_1^\pi$ be given.

Glue $F_1^0\cup (d\circ c)(F_2^\pi)\subset S_1$ to $(b\circ a)(F_1^0)\cup F_2^\pi\subset S_2$ via the above identification, to obtain the required space $X$.

$X$ is clearly a continuum, and is decomposable, because is the union of two proper subcontinua (copies of $S_1,S_2$).

**Edit 1.b:**
The observation motivating the example was that $S_1$ and $S_2$ are separately indecomposable, and an open set $U$ meeting $S_1$ necessarily meets the fiber $F_1^0$ and since $F_1^0$ is nowhere dense in $F_2^0$ it follows that $U$ meets $F_2^\pi$ as well. This property is achieved in a less trivial way than just identifying (an open subset of) a fiber of $S_1$ with a similar subset of $S_2$. Indeed there is no small neighborhood of $F_2^0$ retracting on it. However this does not really help in solving the question.