Hausdorff dimension of inverse images. Let $f: \mathbb{R}^d \to \mathbb{R}$ be a continuous function. Let $t \in (\inf(f), \sup(f))$ and define $C = f^{-1} (t)$. Is it true that the Hausdorff dimension of C is $\geq d -1$? If no how does one construct a counter example?

I believe the following argument works for $d = 2$: 
$A = f^{-1}((-\infty, t))$ and $B= f^{-1}((t,\infty))$ are two open sets whose complement is contained in $C$. If the Hausdorff dimension of $C$ was $< 1$, then $C$ would be totally disconnected. Hence, $\mathbb{R}^2 \setminus C$ would be disconnected, which is implossible.
 A: The boundary of $A = f^{-1}((-\infty, t))$ and $B= f^{-1}((t,\infty))$ is $C = f^{-1} (t)$. Therefore $C$ has Hausdorff dimension at least $d-1$, using this MO entry. I recommend Sergei Ivanov's response for a simple proof. Strictly speaking, the quoted MO entry would require that $A$ or $B$ is bounded, but read below.
Sergei Ivanov's argument can be adapted for a direct proof as follows. One can find balls $A'\subset A$ and $B'\subset B$ of equal radius. Consider the line $L$ connecting the centers of these balls, and the planes orthogonal to $L$ passing through the centers. These planes intersect $A'$ and $B'$ in two parallel disks of dimension $d-1$ and equal radius. Applying the intermediate value theorem to $f$ restricted to the lines parallel to $L$, one sees that the orthogonal projection of $C$ to either disk is surjective, hence $C$ has Hausdorff dimension at least $d-1$.
A: Indeed, more is true:  (1) The topological dimension is $\ge d-1$.  And (2) the Hausdorff dimension is $\ge$ the topological dimension.  
For (1) note that $C$ is a closed set that separates $\mathbb R^d$.
