Let $X_0 \le X$ be a finite-dimensional subspace of $X$ and let $\{e_1, \ldots, e_n\}$ be a basis for $X_0$. Then every $x \in X_0$ has a unique representation as a linear combination of $e_1, \ldots, e_n$:
$$x = \alpha_1(x)e_1 + \ldots \alpha_n(x)e_n$$
Notice that $\alpha_i : X_0 \to \mathbb{F}$ are linear functionals, and they are bounded because since functionals on a finite-dimensional space are always bounded.
Thus, using the Hahn-Banach theorem, we can extend them to bounded functionals $\tilde\alpha_i : X \to \mathbb{F}$.
Now define $P : X \to X_0$ as
$$Px = \tilde\alpha_1(x)e_1 + \ldots \tilde\alpha_n(x)e_n, \quad x\in X$$
Notice that $P$ acts as identity on $X_0$:
$$Px = \tilde\alpha_1(\underbrace{x}_{\in X_0})e_1 + \ldots \tilde\alpha_n(\underbrace{x}_{\in X_0})e_n = \alpha_1(x)e_1 + \ldots \alpha_n(x)e_n = x, \quad x \in X_0$$
Therefore, $\DeclareMathOperator{\Ima}{Im}$$P^2 = P$ since $\Ima P = X_0$.
Finally, $P$ is bounded:
\begin{align}
\|Px\| &= \|\tilde\alpha_1(x)e_1 + \ldots \tilde\alpha_n(x)e_n\| \\
&\le \|\tilde\alpha_1(x)\|\|e_1\| + \ldots \|\tilde\alpha_n(x)\|\|e_n\| \\
&\le \|\tilde\alpha_1\|\|x\|\|e_1\| + \ldots \|\tilde\alpha_n\|\|x\|\|e_n\| \\
&= \big(\|\tilde\alpha_1\|\|e_1\| + \ldots \|\tilde\alpha_n\|\|e_n\|\big)\|x\|
\end{align}
Thus, $P$ is the desired projection.
From $\DeclareMathOperator{\Ker}{Ker}$here we also get the decomposition $X = X_0 \dot+ \Ker P = X_0 \dot+ \bigcap_{i=1}^n \Ker\tilde\alpha_i$. Thus follows that a finite-dimensional subspace always has a direct complement, a fact which cannot be used before conducting a proof similar to this one.
Using some "well known" facts, this is not so difficult to prove.
Also, we assume the coefficient field is $\mathbb R$ or $\mathbb C$. Let's denote this coefficient field by $\mathbb K$.
Our first "well known" fact is, that on a finite dimensional vector space over $\mathbb K$, all norms are equivalent (see, for example, here: http://mathonline.wikidot.com/equivalence-of-norms-in-a-finite-dimensional-linear-space).
One useful consequence of this fact is that in the normed space $Y$ with the norm $||\cdot||$ inherited from $X$, the Heine-Borel theorem (see, for example, here: https://en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem) is true, i. e. a subset $S \subseteq Y$ is compact iff it is $||\cdot||$-closed and $||\cdot||$-bounded.
Now let's fix $x \in X$. Then the function
$$
\varphi: Y \rightarrow \mathbb R_+, \quad y \mapsto ||y - x||
$$
is continuous on $Y$ (see, for example, here: Is the distance function in a metric space (uniformly) continuous?).
Now we will try to find a minimum of $\varphi$. Observe that $\varphi(0) = ||x||$, and if $||y|| > 3 ||x||$ then we have, using the triangle inequality,
$$
\varphi(y) = ||y - x|| \geq ||y|| - ||x|| > 2 ||x||.
$$
If we denote by
$$
\overline{B_Y(0, 3||x||)} = \{ y \in Y | ||y|| \leq 3||x|| \}
$$
the closed $||\cdot||$-ball in $Y$ centered at $0$ with radius $3||x||$, then we can rephrase the preceding two observations as follows. First, if $y \not\in \overline{B_Y(0, 3||x||)}$, then $\varphi(y) > 2 ||x||$. And second, for $0 \in \overline{B_Y(0, 3||x||)}$, we have $\varphi(0) = ||x|| \leq 2 ||x||$ (note that we could have $x = 0$).
This means, that if $\varphi$ attains a minimum on $Y$, this minimum must be attained at a point in $\overline{B_Y(0, 3||x||)}$.
Now, $\overline{B_Y(0, 3||x||)}$ is closed and bounded, hence compact (see above), and $\varphi$ is continuous (see above), and suddenly we recall yet another "well known" fact, namely that a continuous real-valued function on a compact set has a finite infimum and attains this infimum (see, for example, here: https://en.wikipedia.org/wiki/Extreme_value_theorem#Generalization_to_metric_and_topological_spaces).
So we have shown that there is a $y_0 \in \overline{B_Y(0, 3||x||)} \subseteq Y$ such that
$$
||y_0 - x|| = \varphi(y_0) = \inf_{y \in \overline{B_Y(0, 3||x||)}}\varphi(y) = \inf_{y \in Y}\varphi(y) = \inf_{y \in Y}||y - x|| = dist(x, Y),
$$
which is exactly what we wanted to prove.
Best Answer
For $(a)$ take $z=\frac{1}{d}\cdot (x-y)$ where $d=d(x,M)$. Then, $||z||=1$. And, for $y'\in M$ we have \begin{align} ||z-y'||&=\bigl|\bigl|\frac{1}{d}(x-y)-y'\bigl|\bigl|\\ &=\frac{1}{d}\bigl|\bigl|x-y-dy'\bigr|\bigr|\\ &\geq \frac{d}{d}=1 \end{align}
Hence, $d(z,M)\geq 1$. For the other inequality, $$d(z,M)\leq ||z-0||=||z||=1$$
Therefore, $d(z,M)=1=||z||$. Now, for $(b)$, pick $x_1$ on the unit sphere. And let $M_1=span(x_1)$. Since, $X$ is infinite dimensional, $M_1\neq X$. Hence, there is some $x\notin M_1$. Since, $M_1$ is a finite dimensional subspace by $(a)$ there is some $x_2$ on the unit sphere such that $d(x_2,M)=1=||x_2||$. Now, let $M_2=span(x_1,x_2)$ and repeat the same argument as before. Continuing this way, recursively, we construct a sequence $x_n$ such that $||x_n-x_m||\geq 1$ for every $n \neq m$.