You can use the following Lorentz Speed Transformation (derivation below) :
\begin{equation}
\left|\mathbf{\dot{r'}}\right|^{2} = c^{2} - \frac{(c^{2} - |\mathbf{\dot{r}}|^2)}{\gamma^{2}(1 - v\dot{x}/c^{2})^{2}} \tag{1}\label{eq:speed-transform}
\end{equation}
which applies to a light wave or a particle seen from two Inertial Frames of Reference (IFR's) in the 'standard setup' - ie an IFR $S'$ moving along the common $x/x'$ direction wrt an IFR $S$, at a velocity $v$ (where $v$ is +ve or -ve). The instantaneous velocity of the light wave/particle is $\mathbf{\dot{r}}$ in S, and $\mathbf{\dot{r'}}$ in $S'$. (Note the dot means differentiation wrt $t$ in S and differentiation wrt $t'$ in $S'$).
This formula proves the conservation of the speed $c$ for any direction of light travel, whereas with the Lorentz Velocity Transformation alone we can only prove it for directions of travel along the x, y, z axes :
\begin{eqnarray}
\dot{x'}& = & \frac{\dot{x} - v}{1 - v\dot{x}/c^{2}} \nonumber \\
\dot{y'} & = & \frac{\dot{y}}{\gamma(1 - v\dot{x}/c^{2})} \nonumber \\
\dot{z'} & = & \frac{\dot{z}}{\gamma(1 - v\dot{x}/c^{2})} \tag{2}\label{eq:velocity-transform}
\end{eqnarray}
by putting $\mathbf{\dot{r}} = (\pm c, 0, 0), (0, \pm c, 0), (0, 0, \pm c)$ and then showing $\left|\mathbf{\dot{r'}}\right|^2 = c^{2}$.
Note the term $1 - v\dot{x}/c^{2}$ in the denominator of (\ref{eq:speed-transform}) and (\ref{eq:velocity-transform}) is always positive so long as we are dealing with the motion of a light wave or particle, since otherwise we would have : $1 \leq v\dot{x}/c^{2} \Rightarrow |v| |\dot{x}| / c^{2} \geq 1 \Rightarrow |\mathbf{\dot{r}}| \geq |\dot{x}| \geq (c/|v|) \cdot c > c$ (since $|v| < c$) - which is a contradiction, because $|\mathbf{\dot{r}}| \leq c$.
Although we initially derive the Lorentz Transformation (LT) on the basis that the speed of light is constant and rectilinear within all IFR's we should really back check that the LT formula we obtain possesses that property for ALL light transmissions - ie in all directions, since we only consider certain specific light directions in our original derivation. Moreover we should check that any speed less than $c$ maps to a speed less than $c$ under the LT.
(Various other properties of LT need to be checked also, eg that it reduces to the Galilean Transformation when $v \ll c$, and that 'causality reversal' cannot occur for 'time-like' or 'light-like' event intervals, ie the sign of $\Delta t$ cannot be flipped for such intervals - this latter property is proved by consideration of 'space-time intervals').
Plugging $|\mathbf{\dot{r}}| = c$ into the above formula (\ref{eq:speed-transform}) we obtain $|\mathbf{\dot{r'}}| = c$, and plugging in $|\mathbf{\dot{r}}| < c$ we obtain $|\mathbf{\dot{r'}}| < c$, as required.
Notice the rhs of the above formula (\ref{eq:speed-transform}) is a function of velocity $\mathbf{\dot{r}}$ (with $\dot{x} = \mathbf{\dot{r} \cdot i}$) and not speed and this correlates with the fact that the LT can map two velocities of the same speed to two velocities of different speeds - eg take velocities $(u, 0, 0)$ and $(-u, 0, 0)$ with $u \in (0, v)$ and $v > 0$, and doing the algebra we would find equal speed in $S'$ would require $u = c$, a contradiction. Thus speed in $S'$ cannot just be a function of speed in $S$. For example a uniform circular motion in the $xy$-plane in $S$ has constant speed but does not appear as a constant speed in $S'$ because on rhs of (\ref{eq:speed-transform}) $\dot{x}$ is variable.
The Lorentz Speed Transformation (\ref{eq:speed-transform}) can be derived from the Lorentz Velocity Transformation (\ref{eq:velocity-transform}) as follows. From (\ref{eq:velocity-transform}) we have :
\begin{eqnarray*}
\left|\mathbf{\dot{r'}}\right|^{2}(1 - v\dot{x}/c^{2})^2 & = & (\dot{x} - v)^{2} + (\dot{y}/\gamma)^2 + (\dot{z}/\gamma)^2 \\
& = & |\mathbf{\dot{r}}|^2 + \left(\frac{1}{\gamma^{2}} - 1\right)\dot{y}^{2} + \left(\frac{1}{\gamma^{2}} - 1\right)\dot{z}^{2} - 2v\dot{x} + v^{2} \\
& = & |\mathbf{\dot{r}}|^2 - \frac{v^{2}}{c^{2}}(\dot{y}^{2} + \dot{z}^{2}) - 2v\dot{x} + v^{2}, \\
& & \mbox{(since} \hspace{0.5em} \frac{1}{\gamma^{2}} - 1 = -\frac{ v^{2} }{ c^{2} } \mbox{)} \\
& = & |\mathbf{\dot{r}}|^2 - \frac{v^{2}}{c^{2}}(|\mathbf{\dot{r}}|^2 - \dot{x}^{2}) - 2v\dot{x} + v^{2} \\
& = & \frac{1}{\gamma^{2}}|\mathbf{\dot{r}}|^2 + \frac{v^{2}}{c^{2}}\dot{x}^{2} - 2v\dot{x} + v^{2} \\
& = & \frac{1}{\gamma^{2}}(|\mathbf{\dot{r}}|^2 - c^{2}) + \frac{v^{2}}{c^{2}}\dot{x}^{2} - 2v\dot{x} + \left(v^{2} + \frac{c^{2}}{\gamma^{2}}\right) \\
& = & \frac{1}{\gamma^{2}}(|\mathbf{\dot{r}}|^2 - c^{2}) + \frac{v^{2}}{c^{2}}\dot{x}^{2} - 2v\dot{x} + c^{2}, \\
& & \mbox{(since} \hspace{0.5em} v^{2} + \frac{c^{2}}{\gamma^{2}} = v^{2} + c^{2}\left(1 - \frac{v^{2}}{c^{2}}\right) = c^{2} \mbox{)} \\
& = & \frac{1}{\gamma^{2}}(|\mathbf{\dot{r}}|^2 - c^{2}) + \left(\frac{v\dot{x}}{c} - c\right)^{2} \\
& = & \frac{|\mathbf{\dot{r}}|^2 - c^{2}}{\gamma^{2}} + c^{2}\left(1 - \frac{v\dot{x}}{c^{2}}\right)^{2} \\
\end{eqnarray*}
from which (\ref{eq:speed-transform}) now follows.
