It is true that $Z$ spans $L$ — even if $X$ isn't ACM. You can also allow $X$ to be singular (but you do need $X$ irreducible and non-degenerate, of course). To illustrate one of the main ideas it is useful to first look at the case when $X$ is a curve.
If $X$ is a curve. Let $M$ be the span of $Z$ and suppose that $M\neq L$. (In the curve case, $L$ will be a hyperplane). Let $p$ be any point of $X$ outside of $Z$ and let $H$ be any hyperplane containing $M$ and $p$. Then $H\cap X$ contains at least $d+1$ points, so by Bezout's theorem the intersection cannot be zero dimensional. Since $X$ is irreducible and one dimensional, this means that the intersection must be all of $X$, so $X$ is contained in $H$, contrary to hypothesis.
The general case. The idea when $k\geqslant 2$ is to show that if $H$ is a general hyperplane containing $L$ then $H \cap X$ is irreducible and non-degenerate (i.e, the intersection $H\cap X$ is not contained in a smaller linear space of $H$). But now all dimensions have been reduced by $1$, and so iterating this procedure reduces us to the curve case, which we've already solved.
To set this up, note that hyperplanes in $\mathbb{P}^n$ containing $L$ are parameterized by a $\mathbb{P}^{k-1}$ (If $V$ is the underlying vector space of $\mathbb{P}^{n}$, $W$ the underlying vector space of $L$, then the hyperplanes are parameterized by the projectivization of $(V/W)^{*}$). We'll use $H$ to refer both to a point of $\mathbb{P}^{k-1}$ and the corresponding hyperplane in $\mathbb{P}^n$ containing $L$. Define $\Gamma\subset \mathbb{P}^{k-1}\times (X\setminus Z)$ to be the set
$$\Gamma = \left\{(H,p) \mid p\in H\right\}$$
i.e, the pairs $(H,p)$ so that $H$ is a hyperplane containing $L$, and $p$ a point of $H\cap X$ not on $Z$.
If we fix $p$, then the set of possible $H$'s satisfying this condition are simply the hyperplanes $H$ containing the span of $L$ and $p$, and this is parameterized by a $\mathbb{P}^{k-2}$. In other words, the fibres of the projection $\Gamma\longrightarrow X\setminus Z$ are $\mathbb{P}^{k-2}$'s. Since $X\setminus Z$ is irreducible, and since $\mathbb{P}^{k-2}$ is irreducible, this implies that $\Gamma$ is irreducible.
Let $\overline{\Gamma}$ be the Zariski-closure of $\Gamma$ in $\mathbb{P}^{k-1}\times X$. Then $\overline{\Gamma}$ is irreducible since $\Gamma$ is. For a fixed $H\in \mathbb{P}^{k-1}$ the fibre of the projection $\overline{\Gamma}\longrightarrow \mathbb{P}^{k-1}$ over $H$ is simply the intersection $X\cap H$, of dimension $k-1$ (and so connected if $k-1\geqslant 1$). Since $\overline{\Gamma}$ is irreducible, and since the fibres are connected, this implies that the generic fibre is also irreducible, i.e, if $H$ is a generic hyperplane containing $L$, then $H\cap X$ is irreducible.
However, the same trick as in the curve case also shows us that for any hyperplane $H$, $H\cap X$ must be non-degenerate. Let $Z=H\cap X$, so that $Z$ is a variety of degree $d$ and dimension $k-1$. Let $M$ be the span of $Z$. If $M\neq H$ then pick any point $p\in X\setminus Z$ and let $H'$ be any hyperplane containing $M$ and $p$. Then $H'\cap X$ can't be all of $X$ (since this would contradict the non-degeneracy of $X$), so $Z'=H'\cap X$ must be a subvariety of dimension $k-1$ (more precisely, all components of $Z'$ have dimension $k-1$) and degree $d$. But $Z$ is therefore a component of $Z'$, and the equality of degrees tells us that $Z'$ can't have any other components so we must have $Z'=Z$. This contradicts the fact that $p\in Z'$ and $p\not\in Z$.
Together this shows the required inductive step: If $H$ is a general hyperplane containing $L$ then $H\cap X$ is irreducible and non-degenerate.
Other remarks. I'm guessing from the setup of the question that you want to apply the result for a particular $L$ that you have chosen. If, in the application, you're allowed to pick a general $L$ then you can say something stronger. The classical uniform position principle (where ''classical'' in this case means ''established by Joe Harris in the 80's'') states that for a general subspace $L$ of dimension $n-k$ the finite set of $d$-points in $Z=L\cap X$ have the property that any subset of $r+1$ of the points (with $r\leqslant n-k$) span a $\mathbb{P}^{r}$. Picking $r=n-k$, this means that any subset of $n-k+1$ points of $Z$ spans all of $L$, and so in particular $Z$ spans $L$. (Note that $d\geqslant n-k+1$; for instance, as a consequence of the argument above: if $d < n-k+1$ then the $d$ points of $Z$ would never span $L$.)