Solve for $a,b,c,d \in \Bbb R$, given that $a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac 25 =0$

Let $F(a,b,c,d) = a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac25$.

With help of a CAS, one can verify

$$\begin{align} F\left(\frac15+p,\frac25+q,\frac35+r,\frac45+s\right) &= p^2 - pq + q^2 - qr + r^2 -rs + s^2\\ &= \frac12\left(p^2 + (p-q)^2 + (q-r)^2 + (r-s)^2 + s^2\right)\end{align}$$

If one set $(a,b,c,d)$ to $\left(\frac15+p,\frac25+q,\frac35+r,\frac45+s\right)$, one find $$\begin{align} F(a,b,c,d) = 0 &\iff p = p-q = q-r = r-s = s = 0\\ &\iff p = q = r = s = 0 \end{align} $$ This implies the equation at hand has a unique solution: $$(a,b,c,d) = \left(\frac15,\frac25,\frac35,\frac45\right)$$

Update

About the question how I come up with this. I first write $F(a,b,c,d)$ as

$$\begin{align} F(a,b,c,d) &= a^2 + b^2 + c^2 + d^2 - ab - bc - cd - da + d(a-1) + \frac25\\ &= \frac12((a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2) + d(a-1) + \frac25 \end{align}\tag{*1} $$ To simplify the term $d(a-1)$, I introduce $\lambda, \mu$ such that

$$\begin{cases} d &= \frac12 + \lambda + \mu\\ a &= \frac12 + \lambda - \mu \end{cases} \quad\implies\quad d(a-1) = \lambda^2 - \left(\frac12+\mu\right)^2 $$ Now $d-a = 2\mu$ and $(a-b)^2 + (b-c)^2 + (c-d)^2 \ge 3\left(\frac{d-a}{3}\right)^2 = \frac43 \mu^2$.

If one substitute this back into $(*1)$, one find

$$F(a,b,c,d) \ge \frac83\mu^2 + \lambda^2 - (\frac12 + \mu)^2 + \frac25 = \lambda^2 + \frac53\left(\mu - \frac{3}{10}\right)^2$$

In order for $F(a,b,c,d) = 0$, we need

$$\lambda = 0,\quad\mu = \frac{3}{10} \quad\text{ and }\quad(a-b)^2 + (b-c)^2 + (c-d)^2 = \frac13(d-a)^2$$ The last condition forces $d-c = c-b = b-a = \frac13(d-a)$ and leads to the solution $(a,b,c,d) = \left(\frac15,\frac25,\frac35,\frac45\right)$. This is a little bit sloppy to describe, so I look at expansion of $F(a,b,c,d)$ near the solution and obtain a simpler description of $F$ in terms of $p,q,r,s$.

Multiply by $2$ and rearrange to \begin{align*}(a-b)^2 + (b-c)^2 + (c-d)^2 + (d-a)^2 + 2ad - 2d + \frac{4}{5} = 0. \tag{$\star$}\end{align*} For fixed $a$ and $d$, the minimum value of $(a-b)^2 + (b-c)^2 + (c-d)^2$ is $\frac{(d - a)^2}{3}$, with equality if and only if $a, b, c, d$ is an arithmetic progression by the lemma below, so the LHS of $(\star)$ is at least $$\frac{4}{3} (d-a)^2 + 2ad - 2d + \frac{4}{5} = \frac{4}{3} \left( a - \frac{d}{4} \right)^2 + \frac{5}{4} \left( d - \frac{4}{5}\right)^2 \tag{$\dagger$}.$$ But $(\dagger)$ is clearly non-negative, and it is zero if and only if $d = 4/5$ and $a = d/4 = 1/5$, but the LHS of $(\star)$ must be zero. From this, $b = 2/5$ and $c = 3/5$ follow, and there can be no other solution.

Lemma. For fixed $x_0$ and $x_n$, the sum $\sum_{i=1}^n (x_i - x_{i-1})^2$ is minimized when the $x_i$ form an arithmetic progression $x_i = \frac{n-i}{n} x_0 + \frac{i}{n} x_n$.

Proof. For $n = 2$, $(x_0 - x_1)^2 + (x_1 - x_2)^2$ can be rearranged as $$ \left( x_1 - \frac{x_0 + x_2}{2} \right)^2 +x_0^2 + x_2^2 - \frac{(x_0 + x_2)^2}{4}. $$ For $n > 2$, if some $x_k$ is not the midpoint of $x_{k-1}$ and $x_{k+1}$, then $(x_k - x_{k-1})^2 + (x_{k+1} - x_k)^2$ can be reduced by moving $x_k$ to the midpoint, leaving the other terms of $\sum_{i=1}^n (x_i - x_{i-1})^2$ alone. So if a minimum exists, it must have evenly spaced $x_i$. And proving that a minimum exists is simple: the possible values of $x_1, \ldots, x_{n-1}$ that can minimize $f(x_1, \ldots, x_{n-1}) = \sum_{i=1}^n (x_i - x_{i-1})^2$ can be bounded in some closed interval $[-R, R]$, and the image of a connected compact set $[-R, R]^n$ under a continuous function $f$ must be compact and connected (that is, a closed bounded interval).

This lemma can be interpreted physically as stating that the potential energy of a chain of $n$ identical springs with unstretched length zero, with the endpoints of the whole chain anchored, is minimized (and thus the forces at each spring endpoint are in equilibrium) when each spring is stretched equally. Here, $x_0$ and $x_n$ are the fixed endpoints, and $x_{i-1}$ and $x_i$ are the endpoints of the $i$th spring.

Your idea of turning the right side into a sum of perfect squares is a good one. Observing that $$ (a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2=2(a^2+b^2+c^2+d^2-ab-bc-cd-da) $$ we multiply $$ a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac{2}{5}=0 $$ by $2$ and rewrite the result as $$ (a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2+2ad-2d+\frac{4}{5}=0. $$ We define $x=a-b$, $y=b-c$, and $z=c-d$, which gives $a=x+y+z+d$. Substituting gives $$ x^2+y^2+z^2+(x+y+z)^2+2d(x+y+z+d-1)+\frac{4}{5}=0, $$ which can be rewritten as $$ x^2+y^2+z^2+(x+y+z+d)^2+d^2-2d+\frac{4}{5}=0 $$ or $$ x^2+y^2+z^2+(x+y+z+d)^2+(d-1)^2=\frac{1}{5}. $$ So a sum of five perfect squares equals $\frac{1}{5}$. We might hope for a solution in which each of the perfect squares is $\left(\pm\frac{1}{5}\right)^2$, and indeed we find that $x=y=z=-\frac{1}{5}$, $d=\frac{4}{5}$ provides such a solution.

Now we ask whether the solution can be perturbed. Letting $x=-\frac{1}{5}+e$, $y=-\frac{1}{5}+f$, $z=-\frac{1}{5}+g$, $d=\frac{4}{5}+h$ and substituting, we get $$ (-1/5+e)^2+(-1/5+f)^2+(-1/5+g)^2+(1/5+e+f+g+h)^2+(-1/5+h)^2=\frac{1}{5}, $$ which simplifies to $$ e^2+f^2+g^2+h^2+(e+f+g+h)^2=0. $$ This forces $e=f=g=h=0$, and therefore the solution is unique.