Counterexample: $\mathbb C\,I\subset B(\ell^2(\mathbb R))$. It is a one-dimensional subalgebra, so not only it is countably decomposable, it is finitely decomposable, even "one-decomposable".
This example is non-degenerate.
Comment: the notation for the von Neumann algebra generated is terrible, as now you have no notation for the union. I will use the union as a union.
Notation: $W^*(S)$ is the von Neumann algebra generated by the set $S$. It is easy to check that $S'=W^*(S)'$: we have $S'=S'''=(S'')'=W^*(S)'$. So you actually only need to do your argument with the union, and not with the algebra generated.
Suppose that $T\in (\bigcup_j M_j)'$. Then, $T\in M_j'$ for any $j$, so $T\in\bigcap_jM_j'$. Conversely, if $T\in \bigcap_jM_j'$ then $T\in M_j'$ for all $j$; so $T\in(\bigcup_jM_j)'$.
Now, using the first equality.
$$
W^*\Big(\bigcup_jM_j\Big)=\Big(\bigcup_jM_j\Big)'' =\Big[\big(\bigcup_jM_j\big)'\Big]'=\Big[\bigcap_jM_j'\Big]'
$$
Edit: proof that the von Neumann algebra generated by a selfadjoint set $S\subset B(H)$ is $S''$.
Since $S''$ is a von Neumann algebra and it contains $S$, then $W^*(S)\subset S''$. Conversely, since $S\subset W^*(S)$, then $W^*(S)'\subset S'$, so $S''\subset W^*(S)''=W^*(S)$.
Best Answer
There are lots of masas in $B(H)$. They can be classified in two kinds, discrete and continuous (and and direct sums thereof).
The canonical example of a continuous masa in your setting would be $L^\infty(\mathbb R)$, seeing as multiplication operators.
The canonical example of a discrete masa is the diagonal masa: you fix an orthonormal basis $\{e_n\}$, and consider the corresponding orthogonal projections $\{E_n\}$. Then $$ A=\{\sum_ka_kE_k:\ a\in\ell^\infty(\mathbb N)\} $$ would be the diagonal masa corresponding to the orthonormal basis $\{e_n\}$. Not that you gain anything, but if you want to make this concrete, you can take $\{e_n\}$ to be the Hermite Polynomials. Or you can use a double index and define $$ e_{n,m}=e^{2\pi in(x-m)}\,1_{[m,m+1)},\qquad n,m\in\mathbb Z. $$ This would make $$ (E_{n,m}f)(x)=\langle f,e_{n,m}\rangle\,e_{n,m}=\bigg(\int_m^{m+1}f(t)\,e^{-2\pi i (t-n)}\,dt\bigg)\,e^{2\pi in(x-m)}\,1_{[m,m+1)}. $$ In this case $A$ would consist of the operators $$ (T_af)(x)=\sum_{n\in\mathbb Z}a_{n,m}\,\bigg(\int_m^{m+1}f(t)\,e^{-2\pi i (t-n)}\,dt\bigg)\,e^{2\pi in(x-m)},\qquad x\in[m,m+1), $$ where $a\in\ell^\infty(\mathbb Z^2)$.