I forgot to take cognizance of the fact that $f$ is defined on $\mathbb R^m, m \geq 1$, and not just on $\mathbb R$. We use the following fact :
If $f$ is a convex function on $\mathbb R^m$ then $f$ is continuous on $\mathbb R^m$.
A proof of this fact can be found here. If $m=1$ then a much simpler proof can be given via the so-called "three-slopes-property" that convex functions enjoy on intervals in $\mathbb R$.
Let's see how one can use this particular fact to produce a proof of the assertion, using the weaker assumption of continuity.
Suppose that $f : \mathbb R^M \to \mathbb R$ is convex. Then, $f$ is continuous on $\mathbb R^M$ by the fact we mentioned earlier. Throughout the rest of this proof, the variables $x_n,y$ will be used to denote various points in $\mathbb R^M$, while the variables $r,s$ will be used to denote various points in $\mathbb R$. When we write $(y,s) \in \mathbb R^{M+1}$, this is the point whose first $M$ coordinates are given by those of $y$ (in that order) and whose last coordinate equals $s$. $P,Q$ will be used to represent arbitrary points in $\mathbb R^{M+1}$, whose coordinates will be given by $P_i,Q_i , i =1,\ldots,M+1$.
Let $x \in \mathbb R^M$ and $t\in \mathbb R$ be such that $t < f(x)$ (these will be fixed for the rest of the proof). Let the graph of $f$, $G(f) \subset \mathbb R^{M+1}$ be defined by $$
G(f) = \{(y,f(y)) : y \in \mathbb R^M\}.
$$
In particular, $(x,t)$ doesn't lie on $G(f)$. Let $d_{M+1} : \mathbb R^{M+1}\times \mathbb R^{M+1} \to \mathbb R$ be given by the usual Euclidean distance $$
d_{M+1}(P,Q) = \sqrt{\sum_{i=1}^{M+1} (P_i-Q_i)^2}.
$$
An analogous definition holds for the distance $d_M : \mathbb R^M \times \mathbb R^M \to \mathbb R$.
We write $\overline{B_{M+1}(P,r)} = \{P'\in \mathbb R^{M+1} : d_{M+1}(P,P') \leq r\}$ for the closed ball of radius $r$ around $P$. The definition of $\overline{B_{M}(y,r)}$ is similar but with $d_{M+1}$ replaced by $d_{M}$.
Let $D = \inf\{d((x,t),P) : P \in G(f)\}$ be the infimum of all distances from points in $G(f)$ to the point $(t,x)$. Note that $D$ is finite since $D \leq f(x)-t$. We will show that for some point $P' \in G(f)$, $$
D = d_{M+1}((x,t),P').
$$
Then, $P'$ is the desired point.
Our strategy, which resembles that of the OP (more on that later) is the following :
Identify a compact subset $G'(f) \subset G(f)$ such that for $P \in G(f) \setminus G'(f)$, $d((x,t),P) > D$. We will show that there are two such candidates for $G'(f)$ : one whose compactness is easier to prove, and the other which is more natural but whose compactness follows as a corollary of the computations for the first candidate.
Use the compactness of $G'(f)$ and the continuity of $d$ along with Weierstrass' theorem to find a minimizer $P' \in G'(f)$ and complete the result.
Let us identify the candidates for the subset $G'(f) \subset G(f)$. I state them below :
$G'_1(f) = \{(y,f(y)) : y \in \overline{B_M(x,D+1)}\}$.
$G_2'(f) = G(f) \cap \overline{B_{M+1}((x,t),f(x)-t)}$.
The second matches the thinking of the OP. The first provides an easier-to-argue alternative.
It is clear that $G'_1(f),G'_2(f) \subset G(f)$. We will now prove that for $(y,s) \in G(f) \setminus G'_1(f)$ or $(y,s) \in G(f) \setminus G'_2(f)$, we have $d_{M+1}((x,t),(y,s)) > D$. Indeed,
$$
(y,s) \in G(f) \setminus G'_1(f) \implies d_{M+1}((x,t),(y,s)) > d_M(x,y) > D+1 > D,
$$
and
$$
(y,s) \in G(f) \setminus G'_1(f) \implies (y,s) \notin \overline{B_{M+1}((x,t),f(x)-t)} \\ \implies d_{M+1}((x,t),(y,s)) > f(x)-t > D.
$$
Our next task is to prove that $G'_1(f)$ and $G'_2(f)$ are both compact.
To see that $G'_1(f)$ is compact, note that $\overline{B_M(x,D+1)}$ is compact. The function $g : \overline{B_M(x,D+1)} \to \mathbb R^{M+1}$ given by $g(y) = (y,f(y))$ is a function whose components are continuous, hence $g$ is continuous. Now, by definition, $G'_1(f) = g(\overline{B_M(x,D+1)})$ is the image of a compact set under a continuous function. Hence, $G'_1(f)$ is a compact set.
To show that $G'_2(f)$ is compact, we first show that $G(f)$ is closed by showing that it contains all its limit points.
Let $(x_n,f(x_n)) \in G(f)$ be such that it converges to some point $(y,s)$ in $\mathbb R^{M+1}$. Then, $x_n \to y$ in $\mathbb R^M$ and $f(x_n) \to s$ in $\mathbb R$ because the components of a convergent sequence also converge. However, $f$ is continuous, hence $f(x_n) \to f(y)$ in $\mathbb R$ as well. Hence, $s= f(y)$ by uniqueness of limits. Thus, $(y,s) = (y,f(y)) \in G(f)$. Since the choice of convergent sequence $(x_n,f(x_n)) \in G(f)$ was arbitrary, $G(f)$ contains all its limit points, hence it's closed.
Now, $\overline{B_{M+1}((x,t),f(x)-t)}$ is a compact set. So $G'_2(f)$ is the intersection of a closed and a compact set, hence it's a compact set itself.
We let $H(f) \subset G(f)$ be equal to either of $G'_1(f)$ or $G'_2(f)$ in what follows. I want to show that from here, the choice doesn't matter. $H(f)$ is compact and if $P \in G(f) \setminus H(f)$, $d_{M+1}((x,t),P)>D$.
The function $d : H(f) \to \mathbb R$ given by $d(p) = d_{M+1}((x,t),P)$ is a continuous function. As $H(f)$ is compact, by the Weierstrass theorem $d$ attains its minimum on the set $H(f)$, so there is a point $P' \in H(f)$ such that $\min_{P \in H(f)} d(P) = d(P')$.
We claim that $D = d(P')$. Indeed, $$
D = \min_{P \in G(f)} d(P) \leq \min_{P \in H(f)} d(P) = d(P')
$$ because a minimum over a larger set results in a smaller number. On the other hand, we have already seen that for every point $P \in G(f) \setminus H(f)$, $d(P)>D$. Hence, $$
D = \min_{P \in G(f)} d(P) = \min_{P \in H(f)} d(P) = d(P')
$$
and $P'$ is the desired point.
Some points :
We never used the fact that $f(x)>t$ , really. We can replace $f(x)-t$ by $|f(x)-t|$ everywhere and obtain the same statement. Essentially, $(x,t)$ is only required to not lie on $G(f)$.
This question is a vast generalization of the above fact. Indeed, in infinite dimensions some of the arguments above break down because closed balls are not compact in infinite dimension, for example. In this case, convexity is strongly used in the proof, which I won't talk about.
