[proofplan]
The strategy is monotonicity. By definition, $E$ contains a unit line segment $S \subseteq E$. The Hausdorff dimension is monotone under inclusion: $S \subseteq E$ implies $\dim_{\mathcal{H}} S \le \dim_{\mathcal{H}} E$. We then compute $\dim_{\mathcal{H}} S = 1$ — the fact that a non-degenerate line segment in any $\mathbb{R}^n$ has Hausdorff dimension exactly $1$ — by exhibiting a positive finite $1$-Hausdorff measure (the Euclidean length) and verifying that the $s$-Hausdorff measure vanishes for $s > 1$ and is infinite for $s < 1$. Combining the two yields $\dim_{\mathcal{H}} E \ge 1$. The proof has two short steps: extract the line segment, and pin down its Hausdorff dimension.
[/proofplan]
[step:Extract a unit line segment from the Kakeya set]
By the definition of a Kakeya set in $\mathbb{R}^n$, for every direction $e \in S^{n-1}$ there exists $a \in \mathbb{R}^n$ such that
\begin{align*}
S(a, e) := \{a + t e : t \in [0, 1]\} \subseteq E.
\end{align*}
Fix any $e_0 \in S^{n-1}$ (for instance $e_0 = e_1$, the first standard basis vector) and the corresponding base point $a_0 \in \mathbb{R}^n$. Define
\begin{align*}
S := S(a_0, e_0) = \{a_0 + t e_0 : t \in [0, 1]\}.
\end{align*}
Then $S$ is a closed line segment of Euclidean length $1$, and $S \subseteq E$.
[/step]
[step:Identify $\dim_{\mathcal{H}} S = 1$ via positive finite $\mathcal{H}^1$-measure]
The line segment $S = \{a_0 + t e_0 : t \in [0, 1]\}$ is the image of the isometric embedding
\begin{align*}
\iota: [0, 1] &\to \mathbb{R}^n \\
t &\mapsto a_0 + t e_0,
\end{align*}
which preserves Euclidean distance: $|\iota(t) - \iota(s)| = |t - s| \cdot |e_0| = |t - s|$ for all $s, t \in [0, 1]$, since $|e_0| = 1$.
[Hausdorff measures](/page/Hausdorff%20Measure) $\mathcal{H}^k$ are invariant under isometric embeddings of metric spaces (this is immediate from the definition of $\mathcal{H}^k$ via covers by sets of bounded diameter — isometries preserve diameters). Therefore for every $k \ge 0$,
\begin{align*}
\mathcal{H}^k(S) = \mathcal{H}^k([0, 1]).
\end{align*}
On $[0, 1] \subset \mathbb{R}$ (with the standard metric), the [$1$-dimensional Hausdorff measure](/page/Hausdorff%20Measure) coincides with [Lebesgue measure](/page/Lebesgue%20Measure): $\mathcal{H}^1([0, 1]) = \mathcal{L}^1([0, 1]) = 1$. So
\begin{align*}
\mathcal{H}^1(S) = 1 \in (0, \infty).
\end{align*}
By the [definition of Hausdorff dimension](/page/Hausdorff%20Dimension) — the unique value $d \ge 0$ such that $\mathcal{H}^s(A) = \infty$ for $s < d$ and $\mathcal{H}^s(A) = 0$ for $s > d$ — and the fact that $\mathcal{H}^k$ takes a finite positive value at exactly one $k$ (or the dimension), we conclude $\dim_{\mathcal{H}} S = 1$.
[/step]
[step:Apply monotonicity of Hausdorff dimension to conclude $\dim_{\mathcal{H}} E \ge 1$]
Hausdorff dimension is monotone under set inclusion: if $A \subseteq B$ then $\dim_{\mathcal{H}} A \le \dim_{\mathcal{H}} B$. This follows from the fact that any cover of $B$ restricts to a cover of $A$, hence $\mathcal{H}^s(A) \le \mathcal{H}^s(B)$ for all $s \ge 0$, and the dimension is the threshold above which the measure vanishes.
By Step 1, $S \subseteq E$. By Step 2, $\dim_{\mathcal{H}} S = 1$. Therefore
\begin{align*}
1 = \dim_{\mathcal{H}} S \le \dim_{\mathcal{H}} E,
\end{align*}
which is the desired bound. This completes the proof.
[/step]
