= Solution
Work in the complete <Banach space> $C([0,1])$ with the <uniform norm>. Each $E_m$ is closed. Indeed if $f_j\to f$ uniformly and $x_j$ witnesses membership in $E_m$, pass to a subsequence with $x_j\to x$. For each $y\in[0,1]$, uniform convergence and continuity give
$$
|f(y)-f(x)|=\lim_j|f_j(y)-f_j(x_j)|\leq m|y-x|^\alpha.
$$
Thus $x$ witnesses membership for $f$.
Each $E_m$ has empty interior. Approximate an arbitrary $f$ uniformly by a <piecewise linear function> $q$ on a uniform mesh, and let $D$ bound the absolute slopes of $q$. On a sufficiently fine uniform refinement of mesh $\delta$, add a triangular oscillation of fixed small amplitude $b$, alternating between $b$ and $-b$ at successive vertices. The resulting function $g$ is still as close to $f$ as desired, and every segment has absolute slope at least $2b/\delta-D$. For any $x$, choose the farther endpoint of a segment containing $x$. Its distance $h$ is between $\delta/2$ and $\delta$, so
$$
\frac{|g(x+h)-g(x)|}{|h|^\alpha}\geq (2b/\delta-D)\delta^{1-\alpha}\min(1,2^{\alpha-1})>m
$$
when $\delta$ is small. Therefore $g\notin E_m$.
By the <Baire category theorem>, the complement of $\bigcup_m E_m$ is dense and nonempty. If the local limsup in the question were finite at some $x$, its nearby differences would satisfy a finite bound. The remaining differences, with $|h|\geq\delta>0$, are bounded by $2\|f\|_\infty\delta^{-\alpha}$. Combining the bounds would put $f$ in some $E_m$, a contradiction. Hence \b[the required limsup is infinite at every point], including one-sided approaches at the endpoints.
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