$\newcommand\ep\varepsilon\newcommand\la\lambda\newcommand\La\Lambda\newcommand\tla{\tilde\lambda}\newcommand\tLa{\tilde\Lambda}\newcommand\ga\gamma\newcommand\Ga\Gamma\newcommand\de\delta$Your conjecture is true, but your reasoning is incorrect.
Let us first show that your conjecture is true. Let $\La$ denote the set of all nondecreasing functions $\la\colon I\to I$ such that $\la(0)=0$ and $\la(1)=1$, where $I:=[0,1]$. Let $\tLa$ denote the set of all continuous strictly increasing functions $\la\in\La$.
Take any $\la\in\La$. The set $D_\la$ of the points of discontinuity of $\la$ is at most countable, and the sum $\sum_{t\in D_\la}J_\la(t)$ of the jumps $J_\la(t):=\la(t+)-\la(t-)$ of the function $\la$ is $\le1$. So, $\la$ can be approximated uniformly by a function in $\La$ with only finitely many points of discontinuity -- see details on this at the end of this answer. So, without loss of generality (wlog) the set $D_\la$ is finite.
Next, the function $\la$ (with only finitely many points of discontinuity) can be approximated uniformly (on $I$) by a strictly increasing function in $\La$ with only finitely many points of discontinuity -- say by the function $\la_h$ for small real $h>0$ such that $\la_h(t)=(\la(t)+ht)/(1+h)$ for $t\in I$.
So, wlog $\la$ is strictly increasing and the set $D_\la$ is finite.
Take some real $\de>0$. For each point $t_*\in D_\la$, consider its $\de$-neighborhood $N_\de(t_*):=I\cap[t_*-\de,t_*+\de]=:[a_{t_*},b_{t_*}]$. Choose $\de$ so that these neighborhoods are pairwise disjoint.
Let the function $\la_\de\in\tLa$ be defined by the following conditions:
- on each interval $[a_{t_*},b_{t_*}]$, $\la_\de$ equals the linear interpolation of $\la$ over this interval;
\begin{equation}
\text{
$\la_\de=\la$ on $I_\de:=I\setminus\bigcup_{t_*\in D_\la}(a_{t_*},b_{t_*})$.} \tag{1}\label{1}
\end{equation}
Take any real $\ep>0$. If $\de$ is small enough, then for each $t_*\in D_\la$ and all $t$ and $s$ in $[a_{t_*},b_{t_*}]$ we have $|X(t)-X(s)|\le\ep$, and hence
$$
\begin{aligned}
&\sup\{|X(t)-Y(\la_\de(t))|\colon t\in[a_{t_*},b_{t_*}]\} \\
&\le\sup\{|X(t)-Y(l)|\colon t\in[a_{t_*},b_{t_*}],
l\in[\la_\de(a_{t_*}),\la_\de(b_{t_*})]\} \\
&=\sup\{|X(t)-Y(l)|\colon t\in[a_{t_*},b_{t_*}],
l\in[\la(a_{t_*}),\la(b_{t_*})]\} \\
&=\sup\{|X(t)-Y(\la(s))|\colon t\in[a_{t_*},b_{t_*}],
s\in[a_{t_*},b_{t_*}]\} \\
&\le\ep+\sup\{|X(s)-Y(\la(s))|\colon
s\in[a_{t_*},b_{t_*}]\}.
\end{aligned}
$$
Similarly (or in particular),
$$
\begin{aligned}
&\sup\{|t-\la_\de(t)|\colon t\in[a_{t_*},b_{t_*}]\} \\
&\le\ep+\sup\{|s-\la(s)|\colon
s\in[a_{t_*},b_{t_*}]\}.
\end{aligned}
$$
Recalling now \eqref{1} and that $\ep>0$ was arbitrary, and letting
$$\rho_\la(X,Y):=\sup_{t\in [0,1]}|t-\la(t)|\vee \sup_{t\in [0,1]}|X(t)-Y(\la(t))|,$$
we see that
$$\inf\{\rho_\la(X,Y)\colon\la\in\tLa\}\le\inf\{\rho_\la(X,Y)\colon\la\in\La\}.$$
The reverse inequality, $\inf\{\rho_\la(X,Y)\colon\la\in\La\}\le\inf\{\rho_\la(X,Y)\colon\la\in\tLa\}$, holds because $\La\supseteq\tLa$. Thus, your conjecture is proved.
Your reasoning was incorrect, because it is not true that
"we could just mollify the function to get some $\gamma_\delta^\star$ where $\gamma_\delta^\star$ is continuous and strictly increasing and $\gamma_\delta^\star \to \gamma$ uniformly on $[0,1]$ as $\delta \downarrow 0$" if $\ga$ is discontinuous: then such an approximation cannot be uniform, because the uniform limit of continuous functions is continuous.
Detail: Take any nondecreasing function $f$ on $[0,1]$ with $f(0)=0$. Let $f_0:=f$. Take then any $t_1$ in the set $D_f$ of points of discontinuity of $f$, with the jump $J_f(t_1)$ of $f$ at $t_1$. Let $f_1:=f_{t_1}$ be the function obtained from $f_0$ by the removal of the jump of $f_0=f$ at $t_1$; that is, let $f_1(t)=f_0(t)$ for $t\in[0,t_1)$ and $f_1(t)=f_0(t)-J_{f_0}(t_1)$ for $t\in[t_1,1]$. Then $f_1$ is nondecreasing, $f_1(0)=0$, $f_1(1)=f_0(1)-J_f(t_1)<f_0(1)$, $f_0-J_f(t_1)\le f_1\le f_0$, $D_{f_1}=D_{f_0}\setminus\{t_1\}$, and $J_{f_1}(t)=J_{f_0}(t)=J_f(t)$ for all $t\in D_{f_1}$.
Continuing thus, we can remove the jumps of $f$ at all points in any finite subset, say $T$, of $D_f$, to get a nondecreasing function $f_T$ on $[0,1]$ such that $f_T(0)=0$, $f_T(1)=1-\sum_{t\in T}J_f(t)$, $f-\sum_{t\in T}J_f(t)\le f_T\le f$, $D_{f_T}=D_{f}\setminus T$, and $J_{f_T}(t)=J_f(t)$ for all $t\in D_{f_T}$. In particular, we have $0=f_T(0)\le f_T(1)=1-\sum_{t\in T}J_f(t)$, so that $\sum_{t\in T}J_f(t)\le1$ for any finite $T\subseteq D_f$. So, $\sum_{t\in D_f}J_f(t)\le1<\infty$ and hence $D_f$ is at most countable, so that we can write
\begin{equation}
D_f=\{t_1,t_2,\dots\}
\end{equation}
for some distinct $t_j$'s in $[0,1]$.
Take now any natural $n$. Let $f_n,f_{n+1},\dots$ be obtained from $f$ by the successive removals of the jumps of $f$ at $t_n,t_{n+1},\dots$. Then $f_n\ge f_{n+1}\ge\cdots$ and hence there is the pointwise limit $g_n:=\lim_{k\to\infty}f_{n+k}$. Then $g_n$ is nondecreasing, $g_n(0)=0$, $g_n(1)=1-\sum_{k\ge0}J_f(t_{n+k})$, $f-\sum_{k\ge0}J_f(t_{n+k})\le g_n\le f$, $D_{g_n}=D_{f}\setminus\{t_n,t_{n+1},\dots\}=\{t_1,\dots,t_{n-1}\}$, and $J_{g_n}(t)=J_f(t)$ for all $t\in D_{g_n}$.
Now take any $\la\in\La$ in place of $f$ above. Take any real $h>0$. Take any natural $n$ such that $\sum_{k\ge0}J_\la(t_{n+k})<h$. Then $\la-h\le g_n\le\la$, so that $g_n$ uniformly approximates $\la$ if $h$ is small enough. In particular, $g_n(1)\ge f(1)-h=1-h>0$ if $h\in(0,1)$. So, for small enough $h>0$, the function $\la_h:=g_n/g_n(1)$ uniformly approximates $\la$. Moreover, $\la_h$ is in $\La$ and the set $D_{\la_h}=\{t_1,\dots,t_{n-1}\}$ is finite.
Thus, indeed, any $\la\in\La$ can be approximated uniformly by a function in $\La$ with only finitely many points of discontinuity.
