YES. In fact “$D\cap I$ is uncountable for every nonempty open interval $I \subseteq [0,1]$” can be strengthened to “$D\cap I$ has Hausdorff dimension 1 for every nonempty open interval $I\subseteq[0,1]$” (thus “uncountable” can be strengthened to “continuum many”, and much more) simultaneously with a strengthening of “Baire class 1” to “semicontinuous”.

This follows from the fact that any $F_{\sigma}$ meager subset of $[0,1]$ can be the discontinuity set of a semicontinuous function $f:[0,1] \rightarrow [0,1]$ (Proof A below) along with the fact that there exists an $F_{\sigma}$ meager subset of $[0,1]$ whose intersection with any nonempty open subinterval of $[0,1]$ has Hausdorff dimension $1$ (Proof B below).

Proof A: Let $D$ be an $F_{\sigma}$ meager subset of $[0,1].$ Express $D$ as the union of a countable (possibly finite) collection $\{P_n\}$ of closed nowhere dense sets and define $f = \sum \left( 2^{-n}\cdot{\chi}_{P_n}\right),$ where ${\chi}_{P_n}$ is the characteristic function of $P_n$ (i.e. ${\chi}_{P_n}(x) = 1$ if $x \in P_{n},$ and ${\chi}_{P_n}(x) = 0$ if $x \notin P_{n}).$ Then $f$ is a bounded non-negative upper semicontinuous function whose discontinuity set is equal to $D.$ Since $f$ is bounded and non-negative, for some $r > 0$ the function $r f$ (has same properties) will have all its values in $[0,1].$

Proof B: Let $\{I_n: \; n=1,2, \ldots\}$ be an enumeration of the rational-endpoint open subintervals of $[0,1]$ and let $C_n$ be a Cantor set (or any closed nowhere dense set) of Hausdorff dimension greater than $1 - \frac{1}{n}$ contained in $I_n.$ Then $D = \cup_{n=1}^{\infty} C_n$ is an $F_{\sigma}$ meager subset of $[0,1]$ whose intersection with any nonempty open subinterval of $[0,1]$ has Hausdorff dimension $1.$ The reason we get “everywhere of Hausdorff dimension 1” is that given any nonempty open interval $I,$ there exist infinitely many $n$ such that $I_n \subseteq I,$ and so for every $\epsilon > 0$ and for every nonempty open interval $I,$ the set $D \cap I$ has Hausdorff dimension greater than $1 - \epsilon,$ which implies that $D \cap I$ has Hausdorff dimension 1.

YES. In fact “$D\cap I$ is uncountable for every nonempty open interval $I \subseteq [0,1]$” can be strengthened to “$D\cap I$ has Hausdorff dimension 1 for every nonempty open interval $I\subseteq[0,1]$” (thus “uncountable” can be strengthened to “continuum many”, and much more) simultaneously with a strengthening of “Baire class 1” to “semicontinuous”.

This follows from the fact that any $F_{\sigma}$ meager subset of $[0,1]$ can be the discontinuity set of a semicontinuous function $f:[0,1] \rightarrow [0,1]$ (Proof A below) along with the fact that there exists an $F_{\sigma}$ meager subset of $[0,1]$ whose intersection with any nonempty open subinterval of $[0,1]$ has Hausdorff dimension $1$ (Proof B below).

Proof A: Let $D$ be an $F_{\sigma}$ meager subset of $[0,1].$ Express $D$ as the union of a countable (possibly finite) collection $\{P_n\}$ of closed nowhere dense sets and define $f = \sum \left( 2^{-n}\cdot{\chi}_{P_n}\right),$ where ${\chi}_{P_n}$ is the characteristic function of $P_n$ (i.e. ${\chi}_{P_n}(x) = 1$ if $x \in P_{n},$ and ${\chi}_{P_n}(x) = 0$ if $x \notin P_{n}).$ Then $f$ is a bounded non-negative upper semicontinuous function whose discontinuity set is equal to $D.$ (This last statement is not intended to be obvious, but a proof is not particularly difficult.) Since $f$ is bounded and non-negative, for some $r > 0$ the function $r f$ (has same properties) will have all its values in $[0,1].$

Proof B: Let $\{I_n: \; n=1,2, \ldots\}$ be an enumeration of the rational-endpoint open subintervals of $[0,1]$ and let $C_n$ be a Cantor set (or any closed nowhere dense set) of Hausdorff dimension greater than $1 - \frac{1}{n}$ contained in $I_n.$ Then $D = \cup_{n=1}^{\infty} C_n$ is an $F_{\sigma}$ meager subset of $[0,1]$ whose intersection with any nonempty open subinterval of $[0,1]$ has Hausdorff dimension $1.$ The reason we get “everywhere of Hausdorff dimension 1” is that given any nonempty open interval $I,$ there exist infinitely many $n$ such that $I_n \subseteq I,$ and so for every $\epsilon > 0$ and for every nonempty open interval $I,$ the set $D \cap I$ has Hausdorff dimension greater than $1 - \epsilon,$ which implies that $D \cap I$ has Hausdorff dimension 1.