The subalgebra $k[x,xy,x{y}^2,\dots ]$ of $k[x,y]$ is not Noetherian

$A$ is a subring of $k[x,y]$. It contains $1=1+x\cdot 0$; it is closed under addition and additive inverses because $k$ and $xk[x,y]$ both are; and $(c+xf)(c^{\mathrm{\prime }}+x{f}^{\mathrm{\prime }})=c{c}^{\mathrm{\prime }}+x(c{f}^{\mathrm{\prime }}+c^{\mathrm{\prime }}f+xf{f}^{\mathrm{\prime }})$ lies in $A$.

$A=k[x,xy,x{y}^2,\dots ]$, the smallest subring of $k[x,y]$ containing $k$ and every $x{y}^i$. That subring is contained in $A$ because $A$ is a subring containing $k$ and each $x{y}^i=0+x\cdot y^i$. Conversely every $c+xf$ lies in it: expanding $f$ as a $k$-linear combination of monomials $x^a{y}^b$ gives $xf$ as a $k$-linear combination of $x^{a+1}{y}^b=x^a\cdot (x{y}^b)$, and each of $x=x{y}^0$ and $x{y}^b$ is one of the listed generators.

The ideal $\mathfrak{a}$ of $A$ generated by $\{x{y}^i:i\in \mathbb{N}\}$ equals $xk[x,y]$. An element of $\mathfrak{a}$ is a finite sum ${\sum }_j{a}_{j}x{y}^{n_j}$ with $a_j\in A$, which lies in $xk[x,y]$; conversely, for $f\in k[x,y]$ expanded as a $k$-linear combination of monomials $x^a{y}^b$, the element $xf$ is the corresponding combination of $x^a\cdot (x{y}^b)$ with $x^a\in A$, hence lies in $\mathfrak{a}$.

Suppose $\mathfrak{a}=(f_1,\dots ,f_m)$ as an ideal of $A$, with $m\in \mathbb{N}$ and $f_j\in \mathfrak{a}$. Since $x=x{y}^0\in \mathfrak{a}$ and $x\mathrm{\ne }0$, the ideal $\mathfrak{a}$ is nonzero, so at least one $f_j$ is nonzero; let $N\in \mathbb{N}$ be an index beyond which every $f_j$ has vanishing $y$-coefficients, that is, the $y^b$-coefficient of every $f_j$ is $0$ for $b>N$. Any element of $(f_1,\dots ,f_m)$ is ${\sum }_j{a}_j{f}_j$ with $a_j\in A$; writing $a_j=c_j+x{h}_j$ with $c_j\in k$ and $h_j\in k[x,y]$, and using $f_j\in xk[x,y]$ from step 2.1, this element equals ${\sum }_j{c}_j{f}_j+x^{2}g$ for some $g\in k[x,y]$. Its $y^{N+1}$-coefficient is therefore the sum of ${\sum }_j{c}_j\cdot 0=0$ and an element of $x^{2}k[x]$, hence lies in $x^{2}k[x]$.

But $x{y}^{N+1}$ lies in $\mathfrak{a}$ and its $y^{N+1}$-coefficient is $x$, which is not in $x^{2}k[x]$: every element of $x^{2}k[x]$ has vanishing coefficient at $x^1$, whereas $x$ has coefficient $1\mathrm{\ne }0$ there. So $x{y}^{N+1}\mathrm{\notin }(f_1,\dots ,f_m)$, contradicting $\mathfrak{a}=(f_1,\dots ,f_m)$. Hence $\mathfrak{a}$ is not finitely generated and $A$ is not Noetherian.

The ambient ring $k[x,y]$ is Noetherian, since $k$ is a field and a polynomial ring in finitely many variables over a Noetherian ring is Noetherian. So the Noetherian ring $k[x,y]$ has the non-Noetherian subring $A$. And $A$ is not of finite type over $k$: an algebra of finite type over the Noetherian ring $k$ would be Noetherian, which $A$ is not, while $k[x,y]$ itself is of finite type over $k$.