Let $U_1$ and $U_2$ be the affine open subsets of $\mathbb{A}^2$ defined by $x\neq0$ and $y\neq0$, respectively. We want to show that $k[U_1\cup U_2] = k[\mathbb{A}^2]$. Since $U_1$ and $U_2$ cover $\mathbb{A}^2\backslash{0}$, it is sufficient to show that $k[U_1\cup U_2] = k[\mathbb{A}^2\backslash{0}]$.

Note that $\mathbb{A}^2\backslash{0} = U_1\cup U_2 \cup (U_1\cap U_2)$. Therefore, we have the following chain of inclusions: \begin{align*} k[\mathbb{A}^2\backslash{0}] &\subseteq k[U_1\cup U_2 \cup (U_1\cap U_2)]\ &\subseteq k[U_1] \cap k[U_2] \cap k[U_1\cap U_2]\ &= k[\mathbb{A}^2]. \end{align*} The first inclusion follows from the fact that $k[\mathbb{A}^2\backslash{0}]$ is the subring of $k(\mathbb{A}^2)$ consisting of rational functions that are regular on $\mathbb{A}^2\backslash{0}$, and every such function is clearly regular on $U_1\cup U_2 \cup (U_1\cap U_2)$. The second inclusion follows from the fact that a rational function that is regular on $U_1\cup U_2 \cup (U_1\cap U_2)$ must be regular on each of these sets individually.

Since we have $k[U_1\cup U_2] = k[\mathbb{A}^2]$, it follows that $\mathbb{A}^2\backslash{0}$ cannot be affine, because an affine variety cannot be covered by two non-empty affine open subsets.

Let $f$ be a homogeneous polynomial of degree $d$ defining $C$. We may assume that $x = [1:0:0]$. Up to projective transformation we may also assume that $L$ is given by $y = 0$. Then $C \cap L$ is defined by the ideal $(f, y)$ in $k[x,y,z]$.

Since $f$ vanishes at $x$, we may write $$f(x,y,z) = z^m g(x,y,z)$$ where $g(x,y,z)$ is a homogeneous polynomial of degree $d-m$ and $g(1,0,0) \neq 0$. Then $$(f, y) = (z^m g(x,y,z), y) = (z^m, y) + (g(x,y,z), y).$$ Since $y$ does not divide $z^m$, we have $(z^m, y) = (y)$, and so $$(f, y) = (y) + (g(x,y,z), y).$$ Thus the intersection multiplicity $M$ is equal to the order of vanishing of $g(x,y,z)$ at $[1:0:0]$, which is at least $m$.

To show that $m = M$ for all but finitely many lines $L$, it suffices to show that the intersection multiplicity $M$ is upper semicontinuous in $L$. In other words, we need to show that if $L_t$ is a family of lines converging to $L$ (in the Zariski topology) then $M_t \geq M$ for $t$ sufficiently close to $0$.

We may assume that the family $L_t$ is given by $y = t$. Let $g_t(x,z)$ be the polynomial of degree $d-m$ defining $C \cap L_t$ at $[1:0:0]$. Then $g_t(x,z) = f(x,t,z)$, and so the coefficient of $z^{d-m}$ in $g_t(x,z)$ is given by $$a_t = f_{0,d-m}(x,t,1).$$ Since $f_{0,d-m}(x,t,1)$ is a polynomial in $x$ and $t$, it is upper semicontinuous in $t$. Therefore, there exists a neighborhood $U$ of $0$ such that $a_t > 0$ for all $t \in U$.

Suppose that $Z$ is not a component of $X$. Then there exists a proper closed subset $Y$ of $X$ such that $Z$ is contained in $Y$. Since $Y$ is a proper subset of $X$, we have $\operatorname{dim}(Y) < \operatorname{dim}(X)$. By the irreducibility of $Z$, we have $Z \not\subseteq Y$. Therefore, there exists a point $p\in Z$ such that $p\notin Y$.

Since $p$ is a point of $Z$, we have $\operatorname{dim}(\mathcal{O}{X,p}) \geq \operatorname{dim}(Z)$, where $\mathcal{O}{X,p}$ is the local ring of $X$ at $p$. On the other hand, since $p\notin Y$, we have $\mathcal{O}{X,p} \subseteq \mathcal{O}{Y,p}$. This implies $\operatorname{dim}(\mathcal{O}{X,p}) \leq \operatorname{dim}(\mathcal{O}{Y,p})$. By the definition of dimension of a local ring, we have $\operatorname{dim}(Y) = \operatorname{dim}(\mathcal{O}_{Y,p})$. Combining these inequalities, we obtain $\operatorname{dim}(Z) \leq \operatorname{dim}(Y)$. This contradicts the assumption that $\operatorname{dim}(Z) = \operatorname{dim}(X)$.

Therefore, our assumption that $Z$ is not a component of $X$ must be false. Thus, $Z$ is a component of $X$.